Term Rewriting System R:
[x, y, z]
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(+(x, y)) -> +'(minus(y), minus(x))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(x)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, minus(y)) -> MINUS(*(x, y))
*'(x, minus(y)) -> *'(x, y)

Furthermore, R contains two SCCs.

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

Dependency Pairs:

MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(y)

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

The following dependency pairs can be strictly oriented:

MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(y)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MINUS(x1)) =  x1 POL(+(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

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

Dependency Pair:

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

*'(x, minus(y)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

The following dependency pair can be strictly oriented:

*'(x, minus(y)) -> *'(x, y)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(*'(x1, x2)) =  x2 POL(minus(x1)) =  1 + x1 POL(+(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polynomial Ordering`

Dependency Pairs:

*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

The following dependency pairs can be strictly oriented:

*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(*'(x1, x2)) =  x2 POL(+(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes