Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHgm.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:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

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:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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

MINUS₂(s₁(X), s₁(Y)) -> P₁(minus₂(X, Y))
DIV₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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:

MINUS₂(s₁(X), s₁(Y)) -> P₁(minus₂(X, Y))
DIV₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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.

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

Used argument filtering: MINUS₂(x1, x2) = x2
s₁(x1) = s₁(x1)
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:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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:

DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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.

DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))

Used argument filtering: DIV₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
p₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

minus₂(X, 0) -> X
minus₂(s₁(X), s₁(Y)) -> p₁(minus₂(X, Y))
p₁(s₁(X)) -> X
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> s₁(div₂(minus₂(X, Y), s₁(Y)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
p₁(s₁(x0))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))

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