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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))
F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`

Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))

Rules:

f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y

The following dependency pairs can be strictly oriented:

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y

Using the Dependency Graph resulted in no new DP problems.

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