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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(f(f(a, x), y), z) -> F(f(x, z), f(y, z))
F(f(f(a, x), y), z) -> F(x, z)
F(f(f(a, x), y), z) -> F(y, z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

F(f(f(a, x), y), z) -> F(y, z)
F(f(f(a, x), y), z) -> F(x, z)
F(f(f(a, x), y), z) -> F(f(x, z), f(y, z))


Rules:


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


Strategy:

innermost




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

F(f(f(a, x), y), z) -> F(x, z)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

F(f(f(a, x), y), z) -> F(y, z)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes