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

Termination of R to be shown. 



   R
     ↳Overlay and local confluence Check



The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost. 


   R
     ↳OC
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Semantic Labelling


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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

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


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SemLab
             ...
               →DP Problem 2
                 ↳Modular Removal of Rules


Dependency Pairs:

F10(a, f00(f11(a, a), x)) -> F10(a, f10(a, x))
F10(a, f00(f11(a, a), x)) -> F10(a, x)


Rules:


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


Strategy:

innermost




We have the following set of usable rules:

f10(a, f01(f11(a, a), x)) -> f00(f11(a, a), f10(a, f11(a, x)))
f10(a, f00(f11(a, a), x)) -> f00(f11(a, a), f10(a, f10(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(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(F_10(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {f11, f00, f10, F10, a}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

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


The result of this processor delivers one new DP problem.



   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SemLab
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules


Dependency Pairs:

F10(a, f00(f11(a, a), x)) -> F10(a, x)
F10(a, f00(f11(a, a), x)) -> F10(a, f10(a, x))


Rule:


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


Strategy:

innermost




We have the following set of usable rules:

f10(a, f00(f11(a, a), x)) -> f00(f11(a, a), f10(a, f10(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(f_11(x1, x2))=  x1 + x2  
  POL(f_00(x1, x2))=  1 + x1 + x2  
  POL(f_10(x1, x2))=  x1 + x2  
  POL(F_10(x1, x2))=  x1 + x2  
  POL(a)=  0  

We have the following set D of usable symbols: {f11, f00, f10, F10, a}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F10(a, f00(f11(a, a), x)) -> F10(a, x)
F10(a, f00(f11(a, a), x)) -> F10(a, f10(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.


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