Term Rewriting System R:
[x, y]
f(x, a(b(y))) -> a(f(a(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(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
         ↳Usable Rules (Innermost)


Dependency Pairs:

F(b(x), y) -> F(x, b(y))
F(a(x), y) -> F(x, a(y))
F(x, a(b(y))) -> F(a(b(x)), y)


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


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


Dependency Pairs:

F(b(x), y) -> F(x, b(y))
F(a(x), y) -> F(x, a(y))
F(x, a(b(y))) -> F(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(x)) -> b(a(x))
There is only one SCC in the graph and, thus, we obtain one new DP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳SRS
             ...
               →DP Problem 3
                 ↳Non-Overlappingness Check


Dependency Pair:

A(b(x)) -> A(x)


Rule:


a(b(x)) -> b(a(x))





R does not overlap into P. Moreover, R is locally confluent (all critical pairs are trivially joinable).Hence we can switch to innermost.
The transformation is resulting in one subcycle: 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳SRS
             ...
               →DP Problem 4
                 ↳Usable Rules (Innermost)


Dependency Pair:

A(b(x)) -> A(x)


Rule:


a(b(x)) -> b(a(x))


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
             ↳SRS
             ...
               →DP Problem 5
                 ↳Size-Change Principle


Dependency Pair:

A(b(x)) -> A(x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. A(b(x)) -> A(x)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
b(x1) -> b(x1)

We obtain no new DP problems.

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