(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