Termination w.r.t. Q proof of TRS/SK904.03.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:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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:

*¹(x, minus(y)) → MINUS(*(x, y))
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
MINUS(+(x, y)) → MINUS(y)
*¹(x, minus(y)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → *¹(x, y)
MINUS(+(x, y)) → +¹(minus(y), minus(x))

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

*¹(x, minus(y)) → MINUS(*(x, y))
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
MINUS(+(x, y)) → MINUS(y)
*¹(x, minus(y)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → *¹(x, y)
MINUS(+(x, y)) → +¹(minus(y), minus(x))

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
*¹(x, minus(y)) → MINUS(*(x, y))
MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, y)
MINUS(+(x, y)) → +¹(minus(y), minus(x))

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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.

MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

*¹(x, minus(y)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, +(y, z)) → *¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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.

*¹(x, +(y, z)) → *¹(x, z)
*¹(x, +(y, z)) → *¹(x, y)

The remaining pairs can at least be oriented weakly.

*¹(x, minus(y)) → *¹(x, y)

Used ordering: Combined order from the following AFS and order.
*¹(x1, x2) = x2
minus(x1) = x1
+(x1, x2) = +(x1, x2)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

*¹(x, minus(y)) → *¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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.

*¹(x, minus(y)) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*¹(x1, x2) = *¹(x2)
minus(x1) = minus(x1)

Recursive Path Order [2].
Precedence:

[*^1₁, minus_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, 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.