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

(0) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

DBL(s(X)) → DBL(X)
DBLS(cons(X, Y)) → DBL(X)
DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
INDX(cons(X, Y), Z) → SEL(X, Z)
INDX(cons(X, Y), Z) → INDX(Y, Z)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(8) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

cons₂: multiset
SEL₁: multiset
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE

(13) Obligation:

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INDX(cons(X, Y), Z) → INDX(Y, Z)

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

Recursive path order with status [RPO].
Precedence:

cons₂ > INDX₁

Status:

cons₂: multiset
INDX₁: multiset

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(s(X)) → DBL(X)

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

s₁ > DBL₁

Status:

DBL₁: multiset
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

DBLS(cons(X, Y)) → DBLS(Y)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBLS(cons(X, Y)) → DBLS(Y)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

cons₂: multiset

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE