(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x
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:
I(+(x, y)) → +1(i(x), i(y))
I(+(x, y)) → I(x)
I(+(x, y)) → I(y)
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, +(y, z)) → +1(x, y)
The TRS R consists of the following rules:
i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x
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 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
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:
i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
I(+(x, y)) → I(y)
I(+(x, y)) → I(x)
The TRS R consists of the following rules:
i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.