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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(a, f(f(a, x), a)) -> F(f(a, f(a, x)), a)
F(a, f(f(a, x), a)) -> F(a, f(a, x))

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Semantic Labelling`

Dependency Pair:

F(a, f(f(a, x), a)) -> F(a, f(a, x))

Rule:

f(a, f(f(a, x), a)) -> f(f(a, f(a, x)), a)

Using Semantic Labelling, the DP problem could be transformed. The following model was found:
 f(x0, x1) =  0 a =  1 F(x0, x1) =  1

From the dependency graph we obtain 1 (labeled) SCCs which each result in correspondingDP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳SemLab`
`           →DP Problem 2`
`             ↳Modular Removal of Rules`

Dependency Pair:

F10(a, f01(f10(a, x), a)) -> F10(a, f10(a, x))

Rules:

f10(a, f01(f10(a, x), a)) -> f01(f10(a, f10(a, x)), a)
f10(a, f01(f11(a, x), a)) -> f01(f10(a, f11(a, x)), a)

We have the following set of usable rules:

f10(a, f01(f10(a, x), a)) -> f01(f10(a, f10(a, x)), a)
f10(a, f01(f11(a, x), a)) -> f01(f10(a, f11(a, x)), a)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(f_11(x1, x2)) =  x1 + x2 POL(f_10(x1, x2)) =  x1 + x2 POL(f_01(x1, x2)) =  1 + x1 + x2 POL(F_10(x1, x2)) =  x1 + x2 POL(a) =  0

We have the following set D of usable symbols: {f11, f10, f01, F10, a}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F10(a, f01(f10(a, x), a)) -> F10(a, f10(a, x))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

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