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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(a, x) -> F(b, f(c, x))
F(a, x) -> F(c, x)
F(a, f(b, x)) -> F(b, f(a, x))
F(a, f(b, x)) -> F(a, x)
F(d, f(c, x)) -> F(d, f(a, x))
F(d, f(c, x)) -> F(a, x)
F(a, f(c, x)) -> F(c, f(a, x))
F(a, f(c, x)) -> F(a, x)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

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


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

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


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))




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