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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

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

Dependency Pairs:

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

Rule:

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

The following dependency pair can be strictly oriented:

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

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

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

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

resulting in one new DP problem.

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

Dependency Pairs:

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

Rule:

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

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

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

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

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

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

Rule:

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

The following dependency pair can be strictly oriented:

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

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

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

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

resulting in one new DP problem.

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

The following remains to be proven:
Dependency Pair:

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

Rule:

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

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