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))

Innermost 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))


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(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))


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))
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))


Strategy:

innermost




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))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:04 minutes