Termination w.r.t. Q proof of TRS/nontermin/CSREx3_3_25

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:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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:

PREFIX₁(L) -> PREFIX₁(L)
ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> ZWADR₂(XS, YS)
PREFIX₁(L) -> ZWADR₂(L, prefix₁(L))
ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> APP₂(Y, cons₂(X, nil))
FROM₁(X) -> FROM₁(s₁(X))
APP₂(cons₂(X, XS), YS) -> APP₂(XS, YS)

The TRS R consists of the following rules:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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:

PREFIX₁(L) -> PREFIX₁(L)
ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> ZWADR₂(XS, YS)
PREFIX₁(L) -> ZWADR₂(L, prefix₁(L))
ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> APP₂(Y, cons₂(X, nil))
FROM₁(X) -> FROM₁(s₁(X))
APP₂(cons₂(X, XS), YS) -> APP₂(XS, YS)

The TRS R consists of the following rules:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

APP₂(cons₂(X, XS), YS) -> APP₂(XS, YS)

The TRS R consists of the following rules:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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.

APP₂(cons₂(X, XS), YS) -> APP₂(XS, YS)

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

POL(APP₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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

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

ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> ZWADR₂(XS, YS)

The TRS R consists of the following rules:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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.

ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> ZWADR₂(XS, YS)

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

POL(ZWADR₂(x₁, x₂)) = 3·x₁ + 3·x₁·x₂ + 3·x₂
POL(cons₂(x₁, x₂)) = 2 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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

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

PREFIX₁(L) -> PREFIX₁(L)

The TRS R consists of the following rules:

app₂(nil, YS) -> YS
app₂(cons₂(X, XS), YS) -> cons₂(X, app₂(XS, YS))
from₁(X) -> cons₂(X, from₁(s₁(X)))
zWadr₂(nil, YS) -> nil
zWadr₂(XS, nil) -> nil
zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(app₂(Y, cons₂(X, nil)), zWadr₂(XS, YS))
prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))

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