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:
0:03 minutes