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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)


Dependency Pair:

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


Rule:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳Semantic Labelling


Dependency Pair:

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


Rule:

none


Strategy:

innermost




Using Semantic Labelling, the DP problem could be transformed. The following model was found:
F(x0, x1)=  1
f(x0, x1)=  1 + x1
a=  0

From the dependency graph we obtain 0 (labeled) SCCs which each result in correspondingDP problem.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes