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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(x, minus(y)) -> MINUS(+(minus(x), y))
+'(x, minus(y)) -> +'(minus(x), y)
+'(x, minus(y)) -> MINUS(x)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
+'(minus(+(x, 1)), 1) -> MINUS(x)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

+'(x, +(y, z)) -> +'(x, y)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(1) =  0 POL(minus(x1)) =  x1 POL(+(x1, x2)) =  1 + x1 + x2 POL(+'(x1, x2)) =  1 + x2

resulting in one new DP problem.

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

Dependency Pair:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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