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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(s(x), y, y) -> F(y, x, s(x))

Furthermore, R contains one SCC.

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

Dependency Pair:

F(s(x), y, y) -> F(y, x, s(x))

Rule:

f(s(x), y, y) -> f(y, x, s(x))

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

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

F(s(x0), s(x'''), s(x''')) -> F(s(x'''), x0, s(x0))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

F(s(x0), s(x'''), s(x''')) -> F(s(x'''), x0, s(x0))

Rule:

f(s(x), y, y) -> f(y, x, s(x))

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

F(s(x0), s(x'''), s(x''')) -> F(s(x'''), x0, s(x0))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pair:

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

Rule:

f(s(x), y, y) -> f(y, x, s(x))

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

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

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

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 Pair:

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

Rule:

f(s(x), y, y) -> f(y, x, s(x))

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

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

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

The transformation is resulting in one new DP problem:

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

Dependency Pair:

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

Rule:

f(s(x), y, y) -> f(y, x, s(x))

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

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

Rule:

f(s(x), y, y) -> f(y, x, s(x))

Termination of R could not be shown.
Duration:
0:00 minutes