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), z)
*(x, x) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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, y)
*1(x, *(y, z)) → *1(*(x, y), z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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, y)
*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(x2)
*(x1, x2)  =  *(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

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

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.