Term Rewriting System R:
[x]
+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Innermost 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) -> +'(y, z)
+'(p1, +(p1, x)) -> +'(p2, x)
+'(p1, +(p2, +(p2, x))) -> +'(p5, x)
+'(p2, p1) -> +'(p1, p2)
+'(p2, +(p1, x)) -> +'(p1, +(p2, x))
+'(p2, +(p1, x)) -> +'(p2, x)
+'(p2, +(p2, p2)) -> +'(p1, p5)
+'(p2, +(p2, +(p2, x))) -> +'(p1, +(p5, x))
+'(p2, +(p2, +(p2, x))) -> +'(p5, x)
+'(p5, p1) -> +'(p1, p5)
+'(p5, +(p1, x)) -> +'(p1, +(p5, x))
+'(p5, +(p1, x)) -> +'(p5, x)
+'(p5, p2) -> +'(p2, p5)
+'(p5, +(p2, x)) -> +'(p2, +(p5, x))
+'(p5, +(p2, x)) -> +'(p5, x)
+'(p5, +(p5, x)) -> +'(p10, x)
+'(p10, p1) -> +'(p1, p10)
+'(p10, +(p1, x)) -> +'(p1, +(p10, x))
+'(p10, +(p1, x)) -> +'(p10, x)
+'(p10, p2) -> +'(p2, p10)
+'(p10, +(p2, x)) -> +'(p2, +(p10, x))
+'(p10, +(p2, x)) -> +'(p10, x)
+'(p10, p5) -> +'(p5, p10)
+'(p10, +(p5, x)) -> +'(p5, +(p10, x))
+'(p10, +(p5, x)) -> +'(p10, x)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`

Dependency Pair:

+'(p10, +(p2, x)) -> +'(p10, x)

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

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

+'(p10, +(p2, x)) -> +'(p10, x)
one new Dependency Pair is created:

+'(p10, +(p2, +(p2, x''))) -> +'(p10, +(p2, x''))

The transformation is resulting in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

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

+'(+(x, y), z) -> +'(x, +(y, z))
no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`

Dependency Pair:

+'(p5, +(p2, x)) -> +'(p5, x)

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

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

+'(p5, +(p2, x)) -> +'(p5, x)
one new Dependency Pair is created:

+'(p5, +(p2, +(p2, x''))) -> +'(p5, +(p2, x''))

The transformation is resulting in no new DP problems.

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