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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
  ↳8 QDP
    ↳9 QDPOrderProof (⇔)
    ↳10 QDP
    ↳11 PisEmptyProof (⇔)
    ↳12 TRUE
  ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE

(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(x2)
s(x1)  =  s
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
cons2 > [SEL1, s]


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)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


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 [RPO].
Precedence:
s1 > DBL1


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: Recursive Path Order [RPO].
Precedence:
trivial


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