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

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
         ↳Forward Instantiation Transformation


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)))





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

F(a, f(x, a)) -> F(f(a, a), f(a, x))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(a, f(x, 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)))





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
         ↳FwdInst
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Polynomial 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)))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule w.r.t. to the implicit AFS 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))=  1 + x1  
  POL(F(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →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)))




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