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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

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

INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
DBL₁(s₁(X)) -> DBL₁(X)
FROM₁(X) -> FROM₁(s₁(X))
INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
DBL₁(s₁(X)) -> DBL₁(X)
FROM₁(X) -> FROM₁(s₁(X))
INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
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.

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SEL₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
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.

INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INDX₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

DBL₁(s₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
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.

DBL₁(s₁(X)) -> DBL₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DBL₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
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.

DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DBLS₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.