(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(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(*(x, y), z) → *1(x, *(y, z))
*1(*(x, y), z) → *1(y, z)
The TRS R consists of the following rules:
*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))
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(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(
x1,
x2) =
x1
*(
x1,
x2) =
*(
x1,
x2)
i(
x1) =
i
1 =
1
0 =
0
Lexicographic path order with status [LPO].
Precedence:
*2 > 1
i > 1
0 > 1
Status:
*2: [1,2]
i: []
1: []
0: []
The following usable rules [FROCOS05] were oriented:
*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))
(4) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))
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