(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)
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), *(x, z)) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(*(x, +(y, z)), u)
+1(*(x, y), +(*(x, z), u)) → +1(y, z)
The TRS R consists of the following rules:
+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)
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 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(+(x, y), z) → +1(x, +(y, z))
+1(*(x, y), *(x, z)) → +1(y, z)
+1(+(x, y), z) → +1(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(y, z)
The TRS R consists of the following rules:
+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.