Termination w.r.t. Q proof of /tmp/tmpV_eefI/Ex6_15_AEL02

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 PisEmptyProof (⇔)

    ↳34 TRUE

  ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 PisEmptyProof (⇔)

    ↳41 TRUE

  ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 PisEmptyProof (⇔)

    ↳50 TRUE

  ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 QDPOrderProof (⇔)

    ↳73 QDP

    ↳74 PisEmptyProof (⇔)

    ↳75 TRUE

  ↳76 QDP

    ↳77 QDPOrderProof (⇔)

    ↳78 QDP

    ↳79 QDPOrderProof (⇔)

    ↳80 QDP

    ↳81 PisEmptyProof (⇔)

    ↳82 TRUE

  ↳83 QDP

    ↳84 QDPOrderProof (⇔)

    ↳85 QDP

    ↳86 QDPOrderProof (⇔)

    ↳87 QDP

    ↳88 QDPOrderProof (⇔)

    ↳89 QDP

    ↳90 PisEmptyProof (⇔)

    ↳91 TRUE

  ↳92 QDP

    ↳93 QDPOrderProof (⇔)

    ↳94 QDP

    ↳95 QDPOrderProof (⇔)

    ↳96 QDP

    ↳97 QDPOrderProof (⇔)

    ↳98 QDP

    ↳99 PisEmptyProof (⇔)

    ↳100 TRUE

  ↳101 QDP

    ↳102 QDPOrderProof (⇔)

    ↳103 QDP

    ↳104 QDPOrderProof (⇔)

    ↳105 QDP

    ↳106 PisEmptyProof (⇔)

    ↳107 TRUE

  ↳108 QDP

    ↳109 QDPOrderProof (⇔)

    ↳110 QDP

    ↳111 QDPOrderProof (⇔)

    ↳112 QDP

    ↳113 QDPOrderProof (⇔)

    ↳114 QDP

    ↳115 PisEmptyProof (⇔)

    ↳116 TRUE

  ↳117 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(sel(0, cons(X, Z))) → MARK(X)
ACTIVE(first(0, Z)) → MARK(nil)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(sel1(s(X), cons(Y, Z))) → MARK(sel1(X, Z))
ACTIVE(sel1(s(X), cons(Y, Z))) → SEL1(X, Z)
ACTIVE(sel1(0, cons(X, Z))) → MARK(quote(X))
ACTIVE(sel1(0, cons(X, Z))) → QUOTE(X)
ACTIVE(first1(0, Z)) → MARK(nil1)
ACTIVE(first1(s(X), cons(Y, Z))) → MARK(cons1(quote(Y), first1(X, Z)))
ACTIVE(first1(s(X), cons(Y, Z))) → CONS1(quote(Y), first1(X, Z))
ACTIVE(first1(s(X), cons(Y, Z))) → QUOTE(Y)
ACTIVE(first1(s(X), cons(Y, Z))) → FIRST1(X, Z)
ACTIVE(quote(0)) → MARK(01)
ACTIVE(quote1(cons(X, Z))) → MARK(cons1(quote(X), quote1(Z)))
ACTIVE(quote1(cons(X, Z))) → CONS1(quote(X), quote1(Z))
ACTIVE(quote1(cons(X, Z))) → QUOTE(X)
ACTIVE(quote1(cons(X, Z))) → QUOTE1(Z)
ACTIVE(quote1(nil)) → MARK(nil1)
ACTIVE(quote(s(X))) → MARK(s1(quote(X)))
ACTIVE(quote(s(X))) → S1(quote(X))
ACTIVE(quote(s(X))) → QUOTE(X)
ACTIVE(quote(sel(X, Z))) → MARK(sel1(X, Z))
ACTIVE(quote(sel(X, Z))) → SEL1(X, Z)
ACTIVE(quote1(first(X, Z))) → MARK(first1(X, Z))
ACTIVE(quote1(first(X, Z))) → FIRST1(X, Z)
ACTIVE(unquote(01)) → MARK(0)
ACTIVE(unquote(s1(X))) → MARK(s(unquote(X)))
ACTIVE(unquote(s1(X))) → S(unquote(X))
ACTIVE(unquote(s1(X))) → UNQUOTE(X)
ACTIVE(unquote1(nil1)) → MARK(nil)
ACTIVE(unquote1(cons1(X, Z))) → MARK(fcons(unquote(X), unquote1(Z)))
ACTIVE(unquote1(cons1(X, Z))) → FCONS(unquote(X), unquote1(Z))
ACTIVE(unquote1(cons1(X, Z))) → UNQUOTE(X)
ACTIVE(unquote1(cons1(X, Z))) → UNQUOTE1(Z)
ACTIVE(fcons(X, Z)) → MARK(cons(X, Z))
ACTIVE(fcons(X, Z)) → CONS(X, Z)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(sel1(X1, X2)) → ACTIVE(sel1(mark(X1), mark(X2)))
MARK(sel1(X1, X2)) → SEL1(mark(X1), mark(X2))
MARK(sel1(X1, X2)) → MARK(X1)
MARK(sel1(X1, X2)) → MARK(X2)
MARK(quote(X)) → ACTIVE(quote(X))
MARK(first1(X1, X2)) → ACTIVE(first1(mark(X1), mark(X2)))
MARK(first1(X1, X2)) → FIRST1(mark(X1), mark(X2))
MARK(first1(X1, X2)) → MARK(X1)
MARK(first1(X1, X2)) → MARK(X2)
MARK(nil1) → ACTIVE(nil1)
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(01) → ACTIVE(01)
MARK(quote1(X)) → ACTIVE(quote1(X))
MARK(s1(X)) → ACTIVE(s1(mark(X)))
MARK(s1(X)) → S1(mark(X))
MARK(s1(X)) → MARK(X)
MARK(unquote(X)) → ACTIVE(unquote(mark(X)))
MARK(unquote(X)) → UNQUOTE(mark(X))
MARK(unquote(X)) → MARK(X)
MARK(unquote1(X)) → ACTIVE(unquote1(mark(X)))
MARK(unquote1(X)) → UNQUOTE1(mark(X))
MARK(unquote1(X)) → MARK(X)
MARK(fcons(X1, X2)) → ACTIVE(fcons(mark(X1), mark(X2)))
MARK(fcons(X1, X2)) → FCONS(mark(X1), mark(X2))
MARK(fcons(X1, X2)) → MARK(X1)
MARK(fcons(X1, X2)) → MARK(X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
SEL1(mark(X1), X2) → SEL1(X1, X2)
SEL1(X1, mark(X2)) → SEL1(X1, X2)
SEL1(active(X1), X2) → SEL1(X1, X2)
SEL1(X1, active(X2)) → SEL1(X1, X2)
QUOTE(mark(X)) → QUOTE(X)
QUOTE(active(X)) → QUOTE(X)
FIRST1(mark(X1), X2) → FIRST1(X1, X2)
FIRST1(X1, mark(X2)) → FIRST1(X1, X2)
FIRST1(active(X1), X2) → FIRST1(X1, X2)
FIRST1(X1, active(X2)) → FIRST1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
QUOTE1(mark(X)) → QUOTE1(X)
QUOTE1(active(X)) → QUOTE1(X)
S1(mark(X)) → S1(X)
S1(active(X)) → S1(X)
UNQUOTE(mark(X)) → UNQUOTE(X)
UNQUOTE(active(X)) → UNQUOTE(X)
UNQUOTE1(mark(X)) → UNQUOTE1(X)
UNQUOTE1(active(X)) → UNQUOTE1(X)
FCONS(mark(X1), X2) → FCONS(X1, X2)
FCONS(X1, mark(X2)) → FCONS(X1, X2)
FCONS(active(X1), X2) → FCONS(X1, X2)
FCONS(X1, active(X2)) → FCONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 15 SCCs with 46 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FCONS(X1, mark(X2)) → FCONS(X1, X2)
FCONS(mark(X1), X2) → FCONS(X1, X2)
FCONS(active(X1), X2) → FCONS(X1, X2)
FCONS(X1, active(X2)) → FCONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FCONS(active(X1), X2) → FCONS(X1, X2)
FCONS(X1, active(X2)) → FCONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FCONS(x1, x2) = FCONS(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > FCONS₂

Status:

FCONS₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FCONS(X1, mark(X2)) → FCONS(X1, X2)
FCONS(mark(X1), X2) → FCONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FCONS(X1, mark(X2)) → FCONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FCONS(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FCONS(mark(X1), X2) → FCONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FCONS(mark(X1), X2) → FCONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FCONS₂

Status:

FCONS₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) TRUE

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

UNQUOTE1(active(X)) → UNQUOTE1(X)
UNQUOTE1(mark(X)) → UNQUOTE1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

UNQUOTE1(active(X)) → UNQUOTE1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1) = UNQUOTE1(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > UNQUOTE1₁

Status:

UNQUOTE1₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

UNQUOTE1(mark(X)) → UNQUOTE1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

UNQUOTE1(mark(X)) → UNQUOTE1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

UNQUOTE(active(X)) → UNQUOTE(X)
UNQUOTE(mark(X)) → UNQUOTE(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

UNQUOTE(active(X)) → UNQUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE(x1) = UNQUOTE(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > UNQUOTE₁

Status:

UNQUOTE₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

UNQUOTE(mark(X)) → UNQUOTE(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

UNQUOTE(mark(X)) → UNQUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(27) TRUE

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S1(active(X)) → S1(X)
S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(active(X)) → S1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S1(x1) = S1(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > S1₁

Status:

S1₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(mark(X)) → S1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S1(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(33) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(34) TRUE

(35) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE1(active(X)) → QUOTE1(X)
QUOTE1(mark(X)) → QUOTE1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE1(active(X)) → QUOTE1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE1(x1) = QUOTE1(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > QUOTE1₁

Status:

QUOTE1₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE1(mark(X)) → QUOTE1(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE1(mark(X)) → QUOTE1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE1(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(40) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(41) TRUE

(42) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > CONS1₂

Status:

CONS1₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(X1, mark(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(mark(X1), X2) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(mark(X1), X2) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS1₂

Status:

CONS1₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(49) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(50) TRUE

(51) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST1(X1, mark(X2)) → FIRST1(X1, X2)
FIRST1(mark(X1), X2) → FIRST1(X1, X2)
FIRST1(active(X1), X2) → FIRST1(X1, X2)
FIRST1(X1, active(X2)) → FIRST1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST1(active(X1), X2) → FIRST1(X1, X2)
FIRST1(X1, active(X2)) → FIRST1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST1(x1, x2) = FIRST1(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > FIRST1₂

Status:

FIRST1₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST1(X1, mark(X2)) → FIRST1(X1, X2)
FIRST1(mark(X1), X2) → FIRST1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST1(X1, mark(X2)) → FIRST1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST1(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST1(mark(X1), X2) → FIRST1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST1(mark(X1), X2) → FIRST1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FIRST1₂

Status:

FIRST1₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(58) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(59) TRUE

(60) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE(active(X)) → QUOTE(X)
QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE(active(X)) → QUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE(x1) = QUOTE(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > QUOTE₁

Status:

QUOTE₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE(mark(X)) → QUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(65) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(66) TRUE

(67) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL1(X1, mark(X2)) → SEL1(X1, X2)
SEL1(mark(X1), X2) → SEL1(X1, X2)
SEL1(active(X1), X2) → SEL1(X1, X2)
SEL1(X1, active(X2)) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(active(X1), X2) → SEL1(X1, X2)
SEL1(X1, active(X2)) → SEL1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL1(x1, x2) = SEL1(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > SEL1₂

Status:

SEL1₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL1(X1, mark(X2)) → SEL1(X1, X2)
SEL1(mark(X1), X2) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(X1, mark(X2)) → SEL1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL1(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL1(mark(X1), X2) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(mark(X1), X2) → SEL1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL1₂

Status:

SEL1₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(73) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(74) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(75) TRUE

(76) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(77) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(active(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > FROM₁

Status:

FROM₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(78) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(79) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(80) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(81) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(82) TRUE

(83) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(84) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = FIRST(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > FIRST₂

Status:

FIRST₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(85) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(86) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(X1, mark(X2)) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(87) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST(mark(X1), X2) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(88) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(mark(X1), X2) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FIRST₂

Status:

FIRST₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(89) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(90) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(91) TRUE

(92) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(93) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > CONS₂

Status:

CONS₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(94) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(95) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(96) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(97) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(98) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(99) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(100) TRUE

(101) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(102) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > S₁

Status:

S₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(103) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(104) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(105) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(106) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(107) TRUE

(108) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(109) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2) = SEL(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

active₁ > SEL₂

Status:

SEL₂: [2,1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(110) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(111) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(X1, mark(X2)) → SEL(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(112) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(113) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(mark(X1), X2) → SEL(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL₂

Status:

SEL₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(114) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(115) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(116) TRUE

(117) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(sel(0, cons(X, Z))) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(sel1(s(X), cons(Y, Z))) → MARK(sel1(X, Z))
MARK(from(X)) → MARK(X)
MARK(sel1(X1, X2)) → ACTIVE(sel1(mark(X1), mark(X2)))
ACTIVE(sel1(0, cons(X, Z))) → MARK(quote(X))
MARK(sel1(X1, X2)) → MARK(X1)
MARK(sel1(X1, X2)) → MARK(X2)
MARK(quote(X)) → ACTIVE(quote(X))
ACTIVE(first1(s(X), cons(Y, Z))) → MARK(cons1(quote(Y), first1(X, Z)))
MARK(first1(X1, X2)) → ACTIVE(first1(mark(X1), mark(X2)))
ACTIVE(quote1(cons(X, Z))) → MARK(cons1(quote(X), quote1(Z)))
MARK(first1(X1, X2)) → MARK(X1)
MARK(first1(X1, X2)) → MARK(X2)
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(quote(s(X))) → MARK(s1(quote(X)))
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(quote1(X)) → ACTIVE(quote1(X))
ACTIVE(quote(sel(X, Z))) → MARK(sel1(X, Z))
MARK(s1(X)) → ACTIVE(s1(mark(X)))
ACTIVE(quote1(first(X, Z))) → MARK(first1(X, Z))
MARK(s1(X)) → MARK(X)
MARK(unquote(X)) → ACTIVE(unquote(mark(X)))
ACTIVE(unquote(s1(X))) → MARK(s(unquote(X)))
MARK(unquote(X)) → MARK(X)
MARK(unquote1(X)) → ACTIVE(unquote1(mark(X)))
ACTIVE(unquote1(cons1(X, Z))) → MARK(fcons(unquote(X), unquote1(Z)))
MARK(unquote1(X)) → MARK(X)
MARK(fcons(X1, X2)) → ACTIVE(fcons(mark(X1), mark(X2)))
ACTIVE(fcons(X, Z)) → MARK(cons(X, Z))
MARK(fcons(X1, X2)) → MARK(X1)
MARK(fcons(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(sel(0, cons(X, Z))) → mark(X)
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(sel1(0, cons(X, Z))) → mark(quote(X))
active(first1(0, Z)) → mark(nil1)
active(first1(s(X), cons(Y, Z))) → mark(cons1(quote(Y), first1(X, Z)))
active(quote(0)) → mark(01)
active(quote1(cons(X, Z))) → mark(cons1(quote(X), quote1(Z)))
active(quote1(nil)) → mark(nil1)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(sel(X, Z))) → mark(sel1(X, Z))
active(quote1(first(X, Z))) → mark(first1(X, Z))
active(unquote(01)) → mark(0)
active(unquote(s1(X))) → mark(s(unquote(X)))
active(unquote1(nil1)) → mark(nil)
active(unquote1(cons1(X, Z))) → mark(fcons(unquote(X), unquote1(Z)))
active(fcons(X, Z)) → mark(cons(X, Z))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(from(X)) → active(from(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(X))
mark(first1(X1, X2)) → active(first1(mark(X1), mark(X2)))
mark(nil1) → active(nil1)
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(01) → active(01)
mark(quote1(X)) → active(quote1(X))
mark(s1(X)) → active(s1(mark(X)))
mark(unquote(X)) → active(unquote(mark(X)))
mark(unquote1(X)) → active(unquote1(mark(X)))
mark(fcons(X1, X2)) → active(fcons(mark(X1), mark(X2)))
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
sel1(mark(X1), X2) → sel1(X1, X2)
sel1(X1, mark(X2)) → sel1(X1, X2)
sel1(active(X1), X2) → sel1(X1, X2)
sel1(X1, active(X2)) → sel1(X1, X2)
quote(mark(X)) → quote(X)
quote(active(X)) → quote(X)
first1(mark(X1), X2) → first1(X1, X2)
first1(X1, mark(X2)) → first1(X1, X2)
first1(active(X1), X2) → first1(X1, X2)
first1(X1, active(X2)) → first1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
quote1(mark(X)) → quote1(X)
quote1(active(X)) → quote1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(X)
unquote(mark(X)) → unquote(X)
unquote(active(X)) → unquote(X)
unquote1(mark(X)) → unquote1(X)
unquote1(active(X)) → unquote1(X)
fcons(mark(X1), X2) → fcons(X1, X2)
fcons(X1, mark(X2)) → fcons(X1, X2)
fcons(active(X1), X2) → fcons(X1, X2)
fcons(X1, active(X2)) → fcons(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.