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


Rules:


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

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


Rule:


a(b(x)) -> b(a(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
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




We have the following set of usable rules:

a(b(x)) -> b(a(a(x)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(b(x1))=  1 + x1  
  POL(a(x1))=  x1  
  POL(A(x1))=  x1  

We have the following set D of usable symbols: {b, a, A}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


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