Termination w.r.t. Q proof of /tmp/tmpBA0vot/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 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 PisEmptyProof (⇔)

    ↳23 TRUE

  ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 PisEmptyProof (⇔)

    ↳37 TRUE

  ↳38 QDP

    ↳39 QDPOrderProof (⇔)

    ↳40 QDP

    ↳41 PisEmptyProof (⇔)

    ↳42 TRUE

  ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 PisEmptyProof (⇔)

    ↳73 TRUE

  ↳74 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(active(X)) → QUOTE(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(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(mark(X)) → QUOTE(X)

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

mark₁ > QUOTE₁

Status:

QUOTE₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

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

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

mark₁ > SEL1₂
active₁ > SEL1₂

Status:

SEL1₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(active(X)) → S1(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

active(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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(mark(X)) → S1(X)

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

mark₁ > S1₁

Status:

S1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

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.

(25) 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)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

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.

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

mark₁ > DBL1₁

Status:

DBL1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

active(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.

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

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.

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(active(X)) → FROM(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(33) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(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.

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

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

mark₁ > FROM₁

Status:

FROM₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(35) 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.

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) Obligation:

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

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.

(39) 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)
INDX(mark(X1), X2) → INDX(X1, X2)
INDX(active(X1), X2) → INDX(X1, X2)
INDX(X1, active(X2)) → INDX(X1, X2)

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

mark₁ > INDX₂
active₁ > INDX₂

Status:

INDX₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(40) Obligation:

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

active(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.

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) Obligation:

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

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.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

mark₁ > SEL₂
active₁ > SEL₂

Status:

SEL₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(45) 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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

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.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

mark₁ > CONS₂
active₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(50) 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.

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) 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.

(54) 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)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

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.

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

mark₁ > DBLS₁

Status:

DBLS₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

active(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) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

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.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(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.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

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

mark₁ > S₁

Status:

S₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(64) 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.

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

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.

(68) 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)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

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

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.

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

mark₁ > DBL₁

Status:

DBL₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(71) 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.

(72) PisEmptyProof (EQUIVALENT transformation)

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

(73) TRUE

(74) 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.