(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, 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(otimes(x, y), z)
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, 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, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
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) =
x2
oplus(
x1,
x2) =
oplus(
x1,
x2)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
trivial
The following usable rules [FROCOS05] were oriented:
none
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, *(y, z)) → *1(otimes(x, 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, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, *(y, z)) → *1(otimes(x, y), z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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(otimes(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) =
x2
*(
x1,
x2) =
*(
x2)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
trivial
The following usable rules [FROCOS05] were oriented:
none
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) TRUE
(12) 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)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, 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(y, z)
*1(+(x, y), z) → *1(x, 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)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
trivial
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, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, 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