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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(0, 1, x) -> F(s(x), x, x)
F(x, y, s(z)) -> F(0, 1, z)

Furthermore, R contains one SCC.

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

Dependency Pairs:

F(x, y, s(z)) -> F(0, 1, z)
F(0, 1, x) -> F(s(x), x, x)

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(x, y, s(z)) -> F(0, 1, z)
two new Dependency Pairs are created:

F(0, 1, s(z'')) -> F(0, 1, z'')
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

F(0, 1, s(z'')) -> F(0, 1, z'')
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')
F(0, 1, x) -> F(s(x), x, x)

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, x) -> F(s(x), x, x)
one new Dependency Pair is created:

F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(0, 1, s(z'')) -> F(0, 1, z'')

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, s(z'')) -> F(0, 1, z'')
two new Dependency Pairs are created:

F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')
three new Dependency Pairs are created:

F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
three new Dependency Pairs are created:

F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

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(0) =  0 POL(1) =  0 POL(s(x1)) =  1 + x1 POL(F(x1, x2, x3)) =  1 + x3

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`

Dependency Pairs:

F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))

Rules:

f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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