Termination w.r.t. Q proof of TRS/Rubiogm.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

  ↳ 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)))

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:

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)))

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:

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)))

Q is empty.
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

  ↳ 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)))

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.

MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

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

POL(MINUS₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ 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)))

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

            ↳ 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)))

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.

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

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

POL(0) = 0
POL(DIV₂(x₁, x₂)) = 2·x₁
POL(minus₂(x₁, x₂)) = x₁
POL(p₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

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

↳ 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)))

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.