Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHCSRSLASHEx3_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.

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 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: APP₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ZWADR₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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.