(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, 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
otimes(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)
oplus(x1, x2)  =  oplus(x2)

Lexicographic Path Order [LPO].
Precedence:
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, 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.

(5) 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)  =  *1(x1, x2)
*(x1, x2)  =  x2
otimes(x1, x2)  =  x1
oplus(x1, x2)  =  oplus(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
[*^12, oplus2]


The following usable rules [FROCOS05] were oriented: none

(6) 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.

(7) 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)
otimes(x1, x2)  =  otimes(x2)

Lexicographic Path Order [LPO].
Precedence:
[*1, otimes1]


The following usable rules [FROCOS05] were oriented: none

(8) 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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE