(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(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)
*1(x, +(y, z)) → +1(*(x, y), *(x, z))
*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
*1(+(x, y), z) → +1(*(x, z), *(y, z))
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(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 2 less nodes.
(4) Complex Obligation (AND)
(5) 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, +(y, z))
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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(y, z)
+1(+(x, y), z) → +1(x, +(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)
0 =
0
i(
x1) =
i
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, +(y, z)) → *1(x, z)
*1(x, +(y, z)) → *1(x, y)
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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, z)
*1(x, +(y, z)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(
x1,
x2) =
x2
+(
x1,
x2) =
+(
x1,
x2)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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, 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)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) TRUE