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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

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

Dependency Pairs:

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

Rule:

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

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

F(f(a, x), a) -> F(f(a, a), f(x, a))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Rule:

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

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

F(f(a, x), a) -> F(f(f(a, a), f(x, a)), a)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Rule:

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

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(a) =  0 POL(f(x1, x2)) =  1 + x2 POL(F(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 4`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

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

Rule:

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

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