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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(x, a(b(c(y)))) -> F(b(c(a(b(x)))), y)
F(a(x), y) -> F(x, a(y))
F(b(x), y) -> F(x, b(y))
F(c(x), y) -> F(x, c(y))

Furthermore, R contains one SCC.

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

Dependency Pairs:

F(c(x), y) -> F(x, c(y))
F(a(x), y) -> F(x, a(y))
F(b(x), y) -> F(x, b(y))
F(x, a(b(c(y)))) -> F(b(c(a(b(x)))), y)

Rules:

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

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

F(a(x), y) -> F(x, a(y))
three new Dependency Pairs are created:

F(a(x''), a(y'')) -> F(x'', a(a(y'')))
F(a(x''), b(y'')) -> F(x'', a(b(y'')))
F(a(x''), c(y'')) -> F(x'', a(c(y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Instantiation Transformation`

Dependency Pairs:

F(a(x''), c(y'')) -> F(x'', a(c(y'')))
F(a(x''), a(y'')) -> F(x'', a(a(y'')))
F(x, a(b(c(y)))) -> F(b(c(a(b(x)))), y)
F(a(x''), b(y'')) -> F(x'', a(b(y'')))
F(b(x), y) -> F(x, b(y))
F(c(x), y) -> F(x, c(y))

Rules:

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

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

F(b(x), y) -> F(x, b(y))
six new Dependency Pairs are created:

F(b(x''), b(y'')) -> F(x'', b(b(y'')))
F(b(c(a(b(x'')))), y'') -> F(c(a(b(x''))), b(y''))
F(b(x''), c(y'')) -> F(x'', b(c(y'')))
F(b(x'), a(a(y''''))) -> F(x', b(a(a(y''''))))
F(b(x'), a(b(y''''))) -> F(x', b(a(b(y''''))))
F(b(x'), a(c(y''''))) -> F(x', b(a(c(y''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Inst`
`             ...`
`               →DP Problem 3`
`                 ↳Instantiation Transformation`

Dependency Pairs:

F(a(x''), a(y'')) -> F(x'', a(a(y'')))
F(b(x'), a(c(y''''))) -> F(x', b(a(c(y''''))))
F(b(x'), a(b(y''''))) -> F(x', b(a(b(y''''))))
F(b(x'), a(a(y''''))) -> F(x', b(a(a(y''''))))
F(b(x''), b(y'')) -> F(x'', b(b(y'')))
F(x, a(b(c(y)))) -> F(b(c(a(b(x)))), y)
F(a(x''), b(y'')) -> F(x'', a(b(y'')))
F(b(x''), c(y'')) -> F(x'', b(c(y'')))
F(b(c(a(b(x'')))), y'') -> F(c(a(b(x''))), b(y''))
F(c(x), y) -> F(x, c(y))
F(a(x''), c(y'')) -> F(x'', a(c(y'')))

Rules:

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

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

F(c(x), y) -> F(x, c(y))
10 new Dependency Pairs are created:

F(c(x''), c(y'')) -> F(x'', c(c(y'')))
F(c(x'), a(a(y''''))) -> F(x', c(a(a(y''''))))
F(c(x'), a(b(y''''))) -> F(x', c(a(b(y''''))))
F(c(x'), a(c(y''''))) -> F(x', c(a(c(y''''))))
F(c(x'), b(b(y''''))) -> F(x', c(b(b(y''''))))
F(c(a(b(x''''))), b(y'''')) -> F(a(b(x'''')), c(b(y'''')))
F(c(x'), b(c(y''''))) -> F(x', c(b(c(y''''))))
F(c(x'), b(a(a(y'''''')))) -> F(x', c(b(a(a(y'''''')))))
F(c(x'), b(a(b(y'''''')))) -> F(x', c(b(a(b(y'''''')))))
F(c(x'), b(a(c(y'''''')))) -> F(x', c(b(a(c(y'''''')))))

The transformation is resulting in one new DP problem:

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

The following remains to be proven:
Dependency Pairs:

F(c(x'), a(a(y''''))) -> F(x', c(a(a(y''''))))
F(c(x'), a(c(y''''))) -> F(x', c(a(c(y''''))))
F(c(x'), b(a(c(y'''''')))) -> F(x', c(b(a(c(y'''''')))))
F(c(x'), a(b(y''''))) -> F(x', c(a(b(y''''))))
F(c(x'), b(a(b(y'''''')))) -> F(x', c(b(a(b(y'''''')))))
F(b(x'), a(b(y''''))) -> F(x', b(a(b(y''''))))
F(c(x'), b(a(a(y'''''')))) -> F(x', c(b(a(a(y'''''')))))
F(b(x'), a(a(y''''))) -> F(x', b(a(a(y''''))))
F(c(x''), c(y'')) -> F(x'', c(c(y'')))
F(c(x'), b(c(y''''))) -> F(x', c(b(c(y''''))))
F(b(x''), c(y'')) -> F(x'', b(c(y'')))
F(c(a(b(x''''))), b(y'''')) -> F(a(b(x'''')), c(b(y'''')))
F(b(x''), b(y'')) -> F(x'', b(b(y'')))
F(x, a(b(c(y)))) -> F(b(c(a(b(x)))), y)
F(a(x''), b(y'')) -> F(x'', a(b(y'')))
F(b(x'), a(c(y''''))) -> F(x', b(a(c(y''''))))
F(a(x''), c(y'')) -> F(x'', a(c(y'')))
F(c(x'), b(b(y''''))) -> F(x', c(b(b(y''''))))
F(b(c(a(b(x'')))), y'') -> F(c(a(b(x''))), b(y''))
F(a(x''), a(y'')) -> F(x'', a(a(y'')))

Rules:

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

Termination of R could not be shown.
Duration:
0:01 minutes