Termination w.r.t. Q proof of /tmp/tmpfJXg_I/Ex5_7_Luc97

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 QDPOrderProof (⇔)

    ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 PisEmptyProof (⇔)

    ↳65 TRUE

  ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 QDPOrderProof (⇔)

    ↳70 QDP

    ↳71 QDPOrderProof (⇔)

    ↳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 QDPOrderProof (⇔)

    ↳88 QDP

    ↳89 PisEmptyProof (⇔)

    ↳90 TRUE

  ↳91 QDP

    ↳92 QDPOrderProof (⇔)

    ↳93 QDP

    ↳94 QDPOrderProof (⇔)

    ↳95 QDP

    ↳96 PisEmptyProof (⇔)

    ↳97 TRUE

  ↳98 QDP

(0) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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(dbl(0)) → MARK(0)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(dbls(nil)) → MARK(nil)
ACTIVE(dbls(cons(X, Y))) → MARK(cons(dbl(X), dbls(Y)))
ACTIVE(dbls(cons(X, Y))) → CONS(dbl(X), dbls(Y))
ACTIVE(dbls(cons(X, Y))) → DBL(X)
ACTIVE(dbls(cons(X, Y))) → DBLS(Y)
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(indx(nil, X)) → MARK(nil)
ACTIVE(indx(cons(X, Y), Z)) → MARK(cons(sel(X, Z), indx(Y, Z)))
ACTIVE(indx(cons(X, Y), Z)) → CONS(sel(X, Z), indx(Y, Z))
ACTIVE(indx(cons(X, Y), Z)) → SEL(X, Z)
ACTIVE(indx(cons(X, Y), Z)) → INDX(Y, 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(dbl1(0)) → MARK(01)
ACTIVE(dbl1(s(X))) → MARK(s1(s1(dbl1(X))))
ACTIVE(dbl1(s(X))) → S1(s1(dbl1(X)))
ACTIVE(dbl1(s(X))) → S1(dbl1(X))
ACTIVE(dbl1(s(X))) → DBL1(X)
ACTIVE(sel1(0, cons(X, Y))) → MARK(X)
ACTIVE(sel1(s(X), cons(Y, Z))) → MARK(sel1(X, Z))
ACTIVE(sel1(s(X), cons(Y, Z))) → SEL1(X, Z)
ACTIVE(quote(0)) → MARK(01)
ACTIVE(quote(s(X))) → MARK(s1(quote(X)))
ACTIVE(quote(s(X))) → S1(quote(X))
ACTIVE(quote(s(X))) → QUOTE(X)
ACTIVE(quote(dbl(X))) → MARK(dbl1(X))
ACTIVE(quote(dbl(X))) → DBL1(X)
ACTIVE(quote(sel(X, Y))) → MARK(sel1(X, Y))
ACTIVE(quote(sel(X, Y))) → SEL1(X, Y)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(dbl(X)) → DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(X))
MARK(dbls(X)) → ACTIVE(dbls(mark(X)))
MARK(dbls(X)) → DBLS(mark(X))
MARK(dbls(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
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(indx(X1, X2)) → ACTIVE(indx(mark(X1), X2))
MARK(indx(X1, X2)) → INDX(mark(X1), X2)
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(X))
MARK(dbl1(X)) → ACTIVE(dbl1(mark(X)))
MARK(dbl1(X)) → DBL1(mark(X))
MARK(dbl1(X)) → MARK(X)
MARK(01) → ACTIVE(01)
MARK(s1(X)) → ACTIVE(s1(mark(X)))
MARK(s1(X)) → S1(mark(X))
MARK(s1(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(mark(X)))
MARK(quote(X)) → QUOTE(mark(X))
MARK(quote(X)) → MARK(X)
DBL(mark(X)) → DBL(X)
DBL(active(X)) → DBL(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
DBLS(mark(X)) → DBLS(X)
DBLS(active(X)) → DBLS(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)
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)
INDX(mark(X1), X2) → INDX(X1, X2)
INDX(X1, mark(X2)) → INDX(X1, X2)
INDX(active(X1), X2) → INDX(X1, X2)
INDX(X1, active(X2)) → INDX(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
DBL1(mark(X)) → DBL1(X)
DBL1(active(X)) → DBL1(X)
S1(mark(X)) → S1(X)
S1(active(X)) → S1(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)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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 12 SCCs with 37 less nodes.

(4) Complex Obligation (AND)

(5) 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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

QUOTE(mark(X)) → QUOTE(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > QUOTE₁

Status:

mark₁: [1]
QUOTE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

QUOTE(active(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

SEL1(X1, mark(X2)) → SEL1(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(mark(X1), X2) → SEL1(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL1₂

Status:

SEL1₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

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

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(X1, active(X2)) → SEL1(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

SEL1(active(X1), X2) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(active(X1), X2) → SEL1(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) 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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(mark(X)) → S1(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > S1₁

Status:

S1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

S1(active(X)) → S1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(26) 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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

DBL1(active(X)) → DBL1(X)
DBL1(mark(X)) → DBL1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL1(mark(X)) → DBL1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL1(x1) = DBL1(x1)
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBL1₁

Status:

mark₁: [1]
DBL1₁: [1]

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

DBL1(active(X)) → DBL1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL1(active(X)) → DBL1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL1(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) 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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > FROM₁

Status:

mark₁: [1]
FROM₁: [1]

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

FROM(active(X)) → FROM(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(40) 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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

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

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INDX(X1, mark(X2)) → INDX(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

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

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INDX(mark(X1), X2) → INDX(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > INDX₂

Status:

INDX₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

INDX(active(X1), X2) → INDX(X1, X2)
INDX(X1, active(X2)) → INDX(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

INDX(X1, active(X2)) → INDX(X1, X2)

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

INDX(active(X1), X2) → INDX(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

INDX(active(X1), X2) → INDX(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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:

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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

SEL(X1, mark(X2)) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(mark(X1), X2) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL₂

Status:

mark₁: [1]
SEL₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(59) Obligation:

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

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

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(X1, active(X2)) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

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

SEL(active(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(active(X1), X2) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(63) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) 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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(68) Obligation:

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

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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(70) Obligation:

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

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

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(71) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(72) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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(active(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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:

DBLS(active(X)) → DBLS(X)
DBLS(mark(X)) → DBLS(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

DBLS(mark(X)) → DBLS(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBLS₁

Status:

mark₁: [1]
DBLS₁: [1]

The following usable rules [FROCOS05] were oriented: none

(79) Obligation:

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

DBLS(active(X)) → DBLS(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

DBLS(active(X)) → DBLS(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [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(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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.

S(mark(X)) → S(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > S₁

Status:

mark₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(86) Obligation:

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

S(active(X)) → S(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(87) 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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(88) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(89) PisEmptyProof (EQUIVALENT transformation)

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

(90) TRUE

(91) Obligation:

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

DBL(active(X)) → DBL(X)
DBL(mark(X)) → DBL(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(92) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(mark(X)) → DBL(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBL₁

Status:

DBL₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(93) Obligation:

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

DBL(active(X)) → DBL(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(94) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(active(X)) → DBL(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(95) Obligation:

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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

(96) PisEmptyProof (EQUIVALENT transformation)

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

(97) TRUE

(98) Obligation:

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

ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(dbls(cons(X, Y))) → MARK(cons(dbl(X), dbls(Y)))
MARK(dbl(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(dbls(X)) → ACTIVE(dbls(mark(X)))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(dbls(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(indx(cons(X, Y), Z)) → MARK(cons(sel(X, Z), indx(Y, Z)))
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(indx(X1, X2)) → ACTIVE(indx(mark(X1), X2))
ACTIVE(dbl1(s(X))) → MARK(s1(s1(dbl1(X))))
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(X))
ACTIVE(sel1(0, cons(X, Y))) → MARK(X)
MARK(dbl1(X)) → ACTIVE(dbl1(mark(X)))
ACTIVE(sel1(s(X), cons(Y, Z))) → MARK(sel1(X, Z))
MARK(dbl1(X)) → MARK(X)
MARK(s1(X)) → ACTIVE(s1(mark(X)))
ACTIVE(quote(s(X))) → MARK(s1(quote(X)))
MARK(s1(X)) → MARK(X)
MARK(sel1(X1, X2)) → ACTIVE(sel1(mark(X1), mark(X2)))
ACTIVE(quote(dbl(X))) → MARK(dbl1(X))
MARK(sel1(X1, X2)) → MARK(X1)
MARK(sel1(X1, X2)) → MARK(X2)
MARK(quote(X)) → ACTIVE(quote(mark(X)))
ACTIVE(quote(sel(X, Y))) → MARK(sel1(X, Y))
MARK(quote(X)) → MARK(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
mark(dbl1(X)) → active(dbl1(mark(X)))
mark(01) → active(01)
mark(s1(X)) → active(s1(mark(X)))
mark(sel1(X1, X2)) → active(sel1(mark(X1), mark(X2)))
mark(quote(X)) → active(quote(mark(X)))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(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)
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)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
dbl1(mark(X)) → dbl1(X)
dbl1(active(X)) → dbl1(X)
s1(mark(X)) → s1(X)
s1(active(X)) → s1(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)

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