Termination w.r.t. Q proof of /tmp/tmpauXK5y/13.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, +(y, z)) → +(*(x, y), *(x, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
*(*(x, y), z) → *(x, *(y, z))
+(+(x, y), z) → +(x, +(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] 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, y)
*¹(+(y, z), x) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
+¹(+(x, y), z) → +¹(y, z)
*¹(*(x, y), z) → *¹(y, z)
+¹(+(x, y), z) → +¹(x, +(y, z))
*¹(+(y, z), x) → +¹(*(x, y), *(x, z))
*¹(*(x, y), z) → *¹(x, *(y, z))
*¹(+(y, z), x) → *¹(x, z)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ 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, y), z) → +(x, +(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

+¹(+(x, y), z) → +¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
+¹(+(x, y), z) → +¹(x, +(y, z))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(*(x₁, x₂)) = x₁ + x₁·x₂ + x₂
POL(*¹(x₁, x₂)) = x₁ + x₁·x₂
POL(+(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

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

The TRS R consists of the following rules:

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

From the DPs we obtained the following set of size-change graphs:

*¹(x, +(y, z)) → *¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(x, +(y, z)) → *¹(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(*(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(*(x, y), z) → *¹(x, *(y, z))
The graph contains the following edges 1 > 1