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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pair:

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


Rule:


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





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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rule:


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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

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


Rule:


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





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented:

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


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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rule:


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





Using the Dependency Graph resulted in no new DP problems.

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