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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) -> A(b(a(b(a(x)))))
A(a(f(x, y))) -> A(b(a(x)))
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> A(b(a(b(a(y)))))
A(a(f(x, y))) -> A(b(a(y)))
A(a(f(x, y))) -> A(y)
F(a(x), a(y)) -> A(f(x, y))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Furthermore, R contains one SCC.

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

Dependency Pairs:

F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
F(a(x), a(y)) -> A(f(x, y))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))

Rules:

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

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

F(a(x), a(y)) -> A(f(x, y))
two new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(b(x'')), a(b(y''))) -> A(b(f(x'', y'')))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Rules:

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

The following dependency pairs can be strictly oriented:

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

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

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 3`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Rules:

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

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