Term Rewriting System R:
[x]
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

SUM(s(x)) -> SUM(x)
SUM1(s(x)) -> SUM1(x)

Furthermore, R contains two SCCs.

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

Dependency Pair:

SUM(s(x)) -> SUM(x)

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

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

SUM(s(x)) -> SUM(x)
one new Dependency Pair is created:

SUM(s(s(x''))) -> SUM(s(x''))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

SUM(s(s(x''))) -> SUM(s(x''))

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

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

SUM(s(s(x''))) -> SUM(s(x''))
one new Dependency Pair is created:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(SUM(x1)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

SUM1(s(x)) -> SUM1(x)

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

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

SUM1(s(x)) -> SUM1(x)
one new Dependency Pair is created:

SUM1(s(s(x''))) -> SUM1(s(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

SUM1(s(s(x''))) -> SUM1(s(x''))

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

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

SUM1(s(s(x''))) -> SUM1(s(x''))
one new Dependency Pair is created:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(SUM1(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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