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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

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

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

Rules:

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

Strategy:

innermost

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))
four new Dependency Pairs are created:

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

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

Rules:

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

Strategy:

innermost

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

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

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

Rules:

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

Strategy:

innermost

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