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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(x, c(x), c(y)) -> F(y, y, f(y, x, y))
F(x, c(x), c(y)) -> F(y, x, y)
F(s(x), y, z) -> F(x, s(c(y)), c(z))

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Instantiation Transformation`
`       →DP Problem 2`
`         ↳Remaining`

Dependency Pair:

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

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

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

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

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

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

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

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

The following dependency pair can be strictly oriented:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  0 POL(s(x1)) =  1 + x1 POL(F(x1, x2, x3)) =  x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Using the Dependency Graph resulted in no new DP problems.

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

The following remains to be proven:
Dependency Pairs:

F(x, c(x), c(y)) -> F(y, x, y)
F(x, c(x), c(y)) -> F(y, y, f(y, x, y))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

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