Term Rewriting System R:
[x, g, h, xs]
ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil))
ap(ap(ap(foldr, g), h), nil) -> h
ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), xs))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AP(ap(f, x), x) -> AP(ap(x, ap(f, x)), ap(ap(cons, x), nil))
AP(ap(f, x), x) -> AP(x, ap(f, x))
AP(ap(f, x), x) -> AP(ap(cons, x), nil)
AP(ap(f, x), x) -> AP(cons, x)
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(g, x), ap(ap(ap(foldr, g), h), xs))
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(g, x)
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(ap(foldr, g), h), xs)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(ap(foldr, g), h), xs)
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(g, x)
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(g, x), ap(ap(ap(foldr, g), h), xs))
AP(ap(f, x), x) -> AP(ap(cons, x), nil)
AP(ap(f, x), x) -> AP(x, ap(f, x))
AP(ap(f, x), x) -> AP(ap(x, ap(f, x)), ap(ap(cons, x), nil))


Rules:


ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil))
ap(ap(ap(foldr, g), h), nil) -> h
ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), xs))





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

AP(ap(f, x), x) -> AP(ap(cons, x), nil)
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



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




The following remains to be proven:
Dependency Pairs:

AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(g, x)
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(g, x), ap(ap(ap(foldr, g), h), xs))
AP(ap(f, x), x) -> AP(x, ap(f, x))
AP(ap(f, x), x) -> AP(ap(x, ap(f, x)), ap(ap(cons, x), nil))
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(ap(foldr, g), h), xs)


Rules:


ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil))
ap(ap(ap(foldr, g), h), nil) -> h
ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), xs))




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