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

Recursive path order with status [RPO].
Precedence:
*^11 > otimes
+2 > *2 > otimes
1 > otimes

Status:
*^11: [1]
*2: multiset
otimes: multiset
+2: multiset
1: multiset

The following usable rules [FROCOS05] were oriented:

*(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))

(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, *(y, z)) → *1(otimes(x, y), z)
*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)  =  *(x1, x2)
otimes(x1, x2)  =  x2
oplus(x1, x2)  =  oplus(x1, x2)
1  =  1
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [RPO].
Precedence:
+2 > *2 > oplus2

Status:
*2: [1,2]
oplus2: multiset
1: multiset
+2: multiset

The following usable rules [FROCOS05] were oriented:

*(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))

(6) 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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE