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
         ↳Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳Scc To SRS


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)


Rule:

none


Strategy:

innermost




It has been determined that showing finiteness of this DP problem is equivalent to showing termination of a string rewrite system.
(Re)applying the dependency pair method (incl. the dependency graph) for the following SRS:

a(b(c(x))) -> b(a(c(b(x))))
The graph does not contain any SCC and, thus, we obtain no new DP problems.

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