Term Rewriting System R:
[x]
f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(s(s(x))) -> F(f(s(x)))
F(s(s(x))) -> F(s(x))

Furthermore, R contains one SCC.

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

Dependency Pairs:

F(s(s(x))) -> F(s(x))
F(s(s(x))) -> F(f(s(x)))

Rules:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

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

F(s(s(x))) -> F(f(s(x)))
two new Dependency Pairs are created:

F(s(s(0))) -> F(s(0))
F(s(s(s(x'')))) -> F(f(f(s(x''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Narrowing Transformation`

Dependency Pairs:

F(s(s(s(x'')))) -> F(f(f(s(x''))))
F(s(s(x))) -> F(s(x))

Rules:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

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

F(s(s(s(x'')))) -> F(f(f(s(x''))))
two new Dependency Pairs are created:

F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(x))) -> F(s(x))

Rules:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

The following dependency pairs can be strictly oriented:

F(s(s(s(s(x'))))) -> F(f(f(f(s(x')))))
F(s(s(s(0)))) -> F(f(s(0)))
F(s(s(x))) -> F(s(x))

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

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(s(x1)) =  1 + x1 POL(f(x1)) =  1 POL(F(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

f(0) -> s(0)
f(s(0)) -> s(0)
f(s(s(x))) -> f(f(s(x)))

Using the Dependency Graph resulted in no new DP problems.

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