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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

+(*(x, y), *(a, y)) → *(+(x, a), y)
*(*(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 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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), *(a, y)) → *(+(x, a), y)
*(*(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(*(x, y), z) → *1(y, z)
The remaining pairs can at least be oriented weakly.

*1(*(x, y), z) → *1(x, *(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(*1(x1, x2)) = (4)x_1 + (4)x_2   
POL(*(x1, x2)) = 4 + x_1 + x_2   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

*(*(x, y), z) → *(x, *(y, z))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

+(*(x, y), *(a, y)) → *(+(x, a), y)
*(*(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(*(x, y), z) → *1(x, *(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(*1(x1, x2)) = (4)x_1   
POL(*(x1, x2)) = 4 + (2)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

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