Termination w.r.t. Q proof of /tmp/tmpRfEddB/Ex3_2_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 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 PisEmptyProof (⇔)

    ↳46 TRUE

  ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 PisEmptyProof (⇔)

    ↳53 TRUE

  ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 QDPOrderProof (⇔)

    ↳58 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))))
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))
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)

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)
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))
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)

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))))
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))
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)

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 8 SCCs with 22 less nodes.

(4) Complex Obligation (AND)

(5) 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))))
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))
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)

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.

FROM(active(X)) → FROM(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[FROM₁, active₁]

Status:

active₁: [1]
FROM₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) 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))))
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))
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)

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.

FROM(mark(X)) → FROM(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

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))))
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))
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)

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:

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))))
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))
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)

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.

INDX(X1, mark(X2)) → INDX(X1, X2)
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].
Quasi-Precedence:

[INDX₂, mark₁]

Status:

INDX₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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))))
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))
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)

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.

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].
Quasi-Precedence:

[INDX₂, active₁]

Status:

active₁: [1]
INDX₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(16) 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))))
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))
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)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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))))
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))
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)

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.

SEL(X1, mark(X2)) → SEL(X1, X2)
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].
Quasi-Precedence:

[SEL₂, mark₁]

Status:

mark₁: [1]
SEL₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(21) 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))))
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))
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)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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].
Quasi-Precedence:

[SEL₂, active₁]

Status:

active₁: [1]
SEL₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(23) 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))))
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))
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)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) 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))))
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))
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)

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.

CONS(X1, mark(X2)) → CONS(X1, X2)
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].
Quasi-Precedence:

[CONS₂, mark₁]

Status:

CONS₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) 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))))
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))
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)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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].
Quasi-Precedence:

[CONS₂, active₁]

Status:

active₁: [1]
CONS₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(30) 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))))
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))
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)

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) 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))))
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))
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)

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.

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) = DBLS(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[DBLS₁, active₁]

Status:

active₁: [1]
DBLS₁: [1]

The following usable rules [FROCOS05] were oriented: none

(35) 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))))
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))
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)

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) 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))))
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))
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)

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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))))
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))
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)

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[S₁, active₁]

Status:

active₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(42) 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))))
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))
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)

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) 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))))
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))
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)

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) 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))))
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))
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)

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

(48) 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) = DBL(x1)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[DBL₁, active₁]

Status:

DBL₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(49) 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))))
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))
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)

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

(50) 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) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(51) 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))))
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))
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)

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) 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))
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(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))))
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))
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)

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → ACTIVE(s(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
dbl(x1) = dbl
s(x1) = s
MARK(x1) = MARK
mark(x1) = mark
dbls(x1) = dbls
cons(x1, x2) = cons
sel(x1, x2) = sel
0 = 0
indx(x1, x2) = indx
from(x1) = from
active(x1) = active
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

0 > [active, nil] > [dbl, MARK, dbls, cons, sel, indx, from] > mark > s

Status:

active: []
MARK: []
sel: []
dbl: []
s: []
0: []
from: []
cons: []
mark: []
dbls: []
nil: []
indx: []

The following usable rules [FROCOS05] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
sel(X1, mark(X2)) → sel(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
indx(X1, mark(X2)) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(56) 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)
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))
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(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))))
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))
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)

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
dbl(x1) = dbl
s(x1) = s
MARK(x1) = MARK
mark(x1) = x1
dbls(x1) = dbls
cons(x1, x2) = cons
sel(x1, x2) = sel
0 = 0
indx(x1, x2) = indx
from(x1) = from
active(x1) = active
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

s > [dbl, MARK, dbls, sel, indx, from] > [cons, active]
0 > [dbl, MARK, dbls, sel, indx, from] > [cons, active]
nil > [cons, active]

Status:

from: []
active: []
MARK: []
sel: []
cons: []
dbls: []
dbl: []
s: []
0: []
nil: []
indx: []

The following usable rules [FROCOS05] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
sel(X1, mark(X2)) → sel(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
indx(X1, mark(X2)) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(58) 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)
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)
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))
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(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))))
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))
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)

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