(0) Obligation:

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

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, 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, z)) → +1(+(x, y), z)
+1(x, +(y, z)) → +1(x, y)
+1(*(x, y), +(x, 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, z)) → +(+(x, y), z)
+(*(x, y), +(x, 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) 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(+(x, y), z)
+1(x, +(y, z)) → +1(x, y)
+1(*(x, y), +(x, 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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x2)
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
+2 > +^11

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE