(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(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:
+1(x, minus(y)) → MINUS(+(minus(x), y))
+1(x, minus(y)) → +1(minus(x), y)
+1(x, minus(y)) → MINUS(x)
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, +(y, z)) → +1(x, y)
+1(minus(+(x, 1)), 1) → MINUS(x)
The TRS R consists of the following rules:
minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(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 1 SCC with 3 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, minus(y)) → +1(minus(x), y)
+1(x, +(y, z)) → +1(x, y)
The TRS R consists of the following rules:
minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.