Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/RubioSLASHrevlist.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV₁(cons₂(X, L)) -> REV1₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))
REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV₁(cons₂(X, L)) -> REV1₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))
REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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.

REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

Used argument filtering: REV1₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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.

REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))

Used argument filtering: REV2₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
REV₁(x1) = x1
rev₁(x1) = x1
rev2₂(x1, x2) = x2
nil = nil
rev1₂(x1, x2) = rev1
0 = 0
s₁(x1) = s
Used ordering: Quasi Precedence: [rev1, 0, s]

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

The set Q consists of the following terms:

rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev₁(nil)
rev₁(cons₂(x0, x1))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.