Termination w.r.t. Q proof of TRS/SK902.22.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:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, 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:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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:

EXP(x, s(y)) → *¹(x, exp(x, y))
*¹(s(x), y) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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:

EXP(x, s(y)) → *¹(x, exp(x, y))
*¹(s(x), y) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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:

EXP(x, s(y)) → *¹(x, exp(x, y))
*¹(s(x), y) → *¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

↳ 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:

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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), s(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)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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) = x1
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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:

EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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.

EXP(x, s(y)) → EXP(x, y)

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

Lexicographic Path Order [19].
Precedence:

s₁ > EXP₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

exp(x, 0) → s(0)
exp(x, s(y)) → *(x, exp(x, y))
*(0, y) → 0
*(s(x), y) → +(y, *(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

exp(x0, 0)
exp(x0, s(x1))
*(0, x0)
*(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(x1))

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