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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Semantic Labelling


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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

From the dependency graph we obtain 1 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳Modular Removal of Rules


Dependency Pairs:

F01(f11(a, a), x) -> F00(f11(x, a), f10(a, f11(a, a)))
F00(f11(a, a), x) -> F01(x, a)


Rules:


f00(f11(a, a), x) -> f00(f01(x, a), f10(a, f11(a, a)))
f01(f11(a, a), x) -> f00(f11(x, a), f10(a, f11(a, a)))


Strategy:

innermost




We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(f_11(x1, x2))=  x1 + x2  
  POL(F_00(x1, x2))=  x1 + x2  
  POL(f_10(x1, x2))=  x1 + x2  
  POL(F_01(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {f11, F00, f10, F01, a}
No Dependency Pairs can be deleted.
2 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 3
                 ↳Unlabel


Dependency Pairs:

F00(f11(a, a), x) -> F01(x, a)
F01(f11(a, a), x) -> F00(f11(x, a), f10(a, f11(a, a)))


Rule:

none


Strategy:

innermost




Removed all semantic labels.

   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 4
                 ↳Semantic Labelling


Dependency Pairs:

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


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 1 (labeled) SCCs which each result in correspondingDP problem.


   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 5
                 ↳Unlabel


Dependency Pair:

F10(f00(a, a), x) -> F10(f00(x, a), f01(a, f00(a, a)))


Rule:

none


Strategy:

innermost




Removed all semantic labels.

   R
     ↳DPs
       →DP Problem 1
         ↳SemLab
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 6
                 ↳Instantiation Transformation


Dependency Pair:

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


Rule:

none


Strategy:

innermost




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

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

F(f(a, a), f(a, f(a, a))) -> F(f(f(a, f(a, a)), a), f(a, f(a, a)))

The transformation is resulting in no new DP problems.


Innermost Termination of R successfully shown.
Duration:
0:01 minutes