Termination Proof using AProVE

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

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

Dependency Pairs:

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

Rules:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Instantiation Transformation

Dependency Pairs:

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

Rules:

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

three new Dependency Pairs are created:

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

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

f(x, a(b(y))) -> f(a(a(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, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

f(x, a(b(y))) -> f(a(a(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, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

f(x, a(b(y))) -> f(a(a(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, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

f(x, a(b(y))) -> f(a(a(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, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 7

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

f(x, a(b(y))) -> f(a(a(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:
0:07 minutes