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))

Innermost 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
Remaining Obligation(s)

The following remains to be proven:
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(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))

Strategy:

innermost

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