Termination w.r.t. Q proof of TRS/nontermin/CSREx3_2

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

↳ QTRS

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

INDX(cons(X, Y), Z) → SEL(X, Z)
DBLS(cons(X, Y)) → DBL(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBLS(cons(X, Y)) → DBLS(Y)
DBL(s(X)) → DBL(X)
FROM(X) → FROM(s(X))
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.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

INDX(cons(X, Y), Z) → SEL(X, Z)
DBLS(cons(X, Y)) → DBL(X)
DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBL(s(X)) → DBL(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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.
We use the reduction pair processor [13].

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.
none
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)

Lexicographic path order with status [19].
Precedence:

s₁ > SEL₁
cons₂ > SEL₁

Status:

SEL₁: [1]
s₁: multiset
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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.
We use the reduction pair processor [13].

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.
none
Used ordering: Combined order from the following AFS and order.
INDX(x1, x2) = INDX(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > INDX₁

Status:

INDX₁: [1]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

DBL(s(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DBL(x1) = DBL(x1)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > DBL₁

Status:

s₁: multiset
DBL₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

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.
none
Used ordering: Combined order from the following AFS and order.
DBLS(x1) = DBLS(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > DBLS₁

Status:

DBLS₁: [1]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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