Termination w.r.t. Q proof of /tmp/tmpLafZOA/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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 PisEmptyProof (⇔)

    ↳23 TRUE

  ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 PisEmptyProof (⇔)

    ↳37 TRUE

  ↳38 QDP

    ↳39 QDPOrderProof (⇔)

    ↳40 QDP

    ↳41 PisEmptyProof (⇔)

    ↳42 TRUE

  ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳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 PisEmptyProof (⇔)

    ↳71 TRUE

  ↳72 QDP

    ↳73 QDPOrderProof (⇔)

    ↳74 QDP

    ↳75 PisEmptyProof (⇔)

    ↳76 TRUE

  ↳77 QDP

    ↳78 QDPOrderProof (⇔)

    ↳79 QDP

    ↳80 QDPOrderProof (⇔)

    ↳81 QDP

    ↳82 PisEmptyProof (⇔)

    ↳83 TRUE

  ↳84 QDP

    ↳85 QDPOrderProof (⇔)

    ↳86 QDP

    ↳87 PisEmptyProof (⇔)

    ↳88 TRUE

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

mark₁ > FCONS₂
active₁ > FCONS₂

Status:

FCONS₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) 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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) 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.

(13) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > UNQUOTE1₁

Status:

UNQUOTE1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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.

(18) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) 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.

(20) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > UNQUOTE₁

Status:

UNQUOTE₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) 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.

(25) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(26) 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.

(27) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > S1₁

Status:

S1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) 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.

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) 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.

(32) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(33) 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.

(34) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > QUOTE1₁

Status:

QUOTE1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(35) 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.

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) 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.

(39) 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)
CONS1(mark(X1), X2) → CONS1(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS1₂
active₁ > CONS1₂

Status:

CONS1₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(40) 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.

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) 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.

(44) 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)
FIRST1(mark(X1), X2) → FIRST1(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FIRST1₂
active₁ > FIRST1₂

Status:

FIRST1₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(45) 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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) 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.

(49) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(50) 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.

(51) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > QUOTE₁

Status:

QUOTE₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(52) 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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) 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.

(56) 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)
SEL1(mark(X1), X2) → SEL1(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL1₂
active₁ > SEL1₂

Status:

SEL1₂: [1,2]
mark₁: [1]
active₁: [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:

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.

(61) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) 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.

(63) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FROM₁

Status:

FROM₁: [1]
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:

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.

(68) 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)
FIRST(mark(X1), X2) → FIRST(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FIRST₂
active₁ > FIRST₂

Status:

FIRST₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) 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.

(70) PisEmptyProof (EQUIVALENT transformation)

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

(71) TRUE

(72) 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.

(73) 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)
CONS(mark(X1), X2) → CONS(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₂
active₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(74) 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.

(75) PisEmptyProof (EQUIVALENT transformation)

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

(76) TRUE

(77) 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.

(78) 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) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(79) 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.

(80) 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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > S₁

Status:

S₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(81) 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.

(82) PisEmptyProof (EQUIVALENT transformation)

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

(83) TRUE

(84) 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.

(85) 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)
SEL(mark(X1), X2) → SEL(X1, X2)
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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL₂
active₁ > SEL₂

Status:

SEL₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(86) 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.

(87) PisEmptyProof (EQUIVALENT transformation)

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

(88) TRUE

(89) 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.