(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(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(2, 2) → .1(1, 0)
*1(3, x) → .1(x, *(min, x))
*1(3, x) → *1(min, x)
*1(2, min) → .1(min, 2)
*1(.(x, y), z) → .1(*(x, z), *(y, z))
*1(.(x, y), z) → *1(x, z)
*1(.(x, y), z) → *1(y, z)
*1(+(y, z), x) → +1(*(x, y), *(x, z))
*1(+(y, z), x) → *1(x, y)
*1(+(y, z), x) → *1(x, z)
+1(x, x) → *1(2, x)
+1(3, x) → .1(1, +(min, x))
+1(3, x) → +1(min, x)
+1(.(x, y), z) → .1(x, +(y, z))
+1(.(x, y), z) → +1(y, z)
+1(*(2, x), x) → *1(3, x)
.1(x, min) → .1(+(min, x), 3)
.1(x, min) → +1(min, x)
.1(x, .(y, z)) → .1(+(x, y), z)
.1(x, .(y, z)) → +1(x, y)
The TRS R consists of the following rules:
*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(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 2 SCCs with 9 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(3, x) → .1(1, +(min, x))
.1(x, .(y, z)) → .1(+(x, y), z)
.1(x, .(y, z)) → +1(x, y)
+1(.(x, y), z) → .1(x, +(y, z))
+1(.(x, y), z) → +1(y, z)
+1(*(2, x), x) → *1(3, x)
*1(3, x) → .1(x, *(min, x))
The TRS R consists of the following rules:
*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)
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:
*1(.(x, y), z) → *1(y, z)
*1(.(x, y), z) → *1(x, z)
*1(+(y, z), x) → *1(x, y)
*1(+(y, z), x) → *1(x, z)
The TRS R consists of the following rules:
*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.