Termination w.r.t. Q proof of TRS/SK902.12.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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), 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:

*¹(s(x), y) → +¹(*(x, y), y)
*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → *¹(x, y)
*¹(p(x), y) → +¹(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
+¹(s(x), y) → +¹(x, y)
+¹(p(x), y) → +¹(x, y)
*¹(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

*¹(s(x), y) → +¹(*(x, y), y)
*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → *¹(x, y)
*¹(p(x), y) → +¹(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
+¹(s(x), y) → +¹(x, y)
+¹(p(x), y) → +¹(x, y)
*¹(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

*¹(s(x), y) → +¹(*(x, y), y)
*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → *¹(x, y)
MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)
*¹(p(x), y) → +¹(*(x, y), minus(y))
+¹(s(x), y) → +¹(x, y)
+¹(p(x), y) → +¹(x, y)
*¹(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(s(x)) → MINUS(x)

The remaining pairs can at least be oriented weakly.

MINUS(p(x)) → MINUS(x)

Used ordering: Combined order from the following AFS and order.
MINUS(x1) = MINUS(x1)
s(x1) = s(x1)
p(x1) = x1

Recursive path order with status [2].
Precedence:

trivial

Status:

MINUS₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

MINUS(p(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(p(x)) → MINUS(x)

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

Recursive path order with status [2].
Precedence:

p₁ > MINUS₁

Status:

MINUS₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

+¹(s(x), y) → +¹(x, y)
+¹(p(x), y) → +¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(p(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.

+¹(s(x), y) → +¹(x, y)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

+^1₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(s(x), 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(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

s₁ > +^1₁

Status:

+^1₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), 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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → *¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(p(x), y) → *¹(x, y)

The remaining pairs can at least be oriented weakly.

*¹(s(x), y) → *¹(x, y)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

*^1₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

*¹(s(x), y) → *¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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(s(x), 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(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

s₁ > *^1₁

Status:

*^1₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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