(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