Termination w.r.t. Q proof of TRS/SK904.09.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
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

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, x2)
+(x1, x2) = +(x1, x2)
i(x1) = i
0 = 0

Lexicographic path order with status [19].
Quasi-Precedence:

[+^1₂, +_2] > 0
i > 0

Status:

+₂: [1,2]
0: multiset
i: []
+^1₂: [1,2]

The following usable rules [14] were oriented:

+(+(x, y), z) → +(x, +(y, z))
+(x, i(x)) → 0
+(x, 0) → x

↳ 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
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

*^1₁: [1]
+₂: [1,2]

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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(y, z)

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, x2)
+(x1, x2) = +(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:

+₂ > *^1₂

Status:

*^1₂: [2,1]
+₂: [2,1]

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
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

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.