Term Rewriting System R:
[x, y, z]
*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

*'(x, *(y, z)) -> *'(*(x, y), z)
*'(x, *(y, z)) -> *'(x, y)
*'(*(i(x), k(y, z)), x) -> K(*(*(i(x), y), x), *(*(i(x), z), x))
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), y), x)
*'(*(i(x), k(y, z)), x) -> *'(i(x), y)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), z), x)
*'(*(i(x), k(y, z)), x) -> *'(i(x), z)
I(*(x, y)) -> *'(i(y), i(x))
I(*(x, y)) -> I(y)
I(*(x, y)) -> I(x)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Remaining`

Dependency Pairs:

I(*(x, y)) -> I(x)
I(*(x, y)) -> I(y)

Rules:

*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1

The following dependency pairs can be strictly oriented:

I(*(x, y)) -> I(x)
I(*(x, y)) -> I(y)

Additionally, the following rules can be oriented:

*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(I(x1)) =  1 + x1 POL(i(x1)) =  x1 POL(1) =  0 POL(*(x1, x2)) =  1 + x1 + x2 POL(k(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Remaining`

Dependency Pair:

Rules:

*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

*'(*(i(x), k(y, z)), x) -> *'(i(x), z)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), z), x)
*'(*(i(x), k(y, z)), x) -> *'(i(x), y)
*'(*(i(x), k(y, z)), x) -> *'(*(i(x), y), x)
*'(x, *(y, z)) -> *'(x, y)
*'(x, *(y, z)) -> *'(*(x, y), z)

Rules:

*(x, 1) -> x
*(1, y) -> y
*(i(x), x) -> 1
*(x, i(x)) -> 1
*(x, *(y, z)) -> *(*(x, y), z)
*(*(x, y), i(y)) -> x
*(*(x, i(y)), y) -> x
*(k(x, y), k(y, x)) -> 1
*(*(i(x), k(y, z)), x) -> k(*(*(i(x), y), x), *(*(i(x), z), x))
i(1) -> 1
i(i(x)) -> x
i(*(x, y)) -> *(i(y), i(x))
k(x, 1) -> 1
k(x, x) -> 1
k(*(x, i(y)), *(y, i(x))) -> 1

Termination of R could not be shown.
Duration:
0:00 minutes