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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

*1(*(x, y), z) → *1(y, z)
+1(*(x, y), *(a, y)) → *1(+(x, a), y)
+1(*(x, y), *(a, y)) → +1(x, a)
*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 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(y, z)
+1(*(x, y), *(a, y)) → *1(+(x, a), y)
+1(*(x, y), *(a, y)) → +1(x, a)
*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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [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)
*(x1, x2)  =  *(x1, x2)

Recursive path order with status [2].
Quasi-Precedence:
*2 > *^11

Status:
*^11: [1]
*2: [1,2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ 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.