(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

+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))
*1(*(x, y), z) → *1(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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(4) Obligation:

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Recursive Path Order [RPO].
Precedence:
[*^11, *2]


The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE