Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/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))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DIV₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = minus₁(x1)
0 = 0
p₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

s₁ > minus₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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.