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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rule for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a)=  1  
  POL(f)=  0  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2) -> x1
f(x1, x2) -> f


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Argument Filtering and Ordering


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rule for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a)=  0  
  POL(F(x1, x2))=  x1 + x2  
  POL(f(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2) -> F(x1, x2)
f(x1, x2) -> f(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

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


Rule:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes