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
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(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




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
Forward Instantiation Transformation


Dependency Pairs:

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


Strategy:

innermost




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

AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(g, x)
two new Dependency Pairs are created:

AP(ap(ap(foldr, ap(ap(foldr, g''), h'')), h), ap(ap(cons, ap(ap(cons, x''), xs'')), xs)) -> AP(ap(ap(foldr, g''), h''), ap(ap(cons, x''), xs''))
AP(ap(ap(foldr, ap(f, x''')), h), ap(ap(cons, x0), xs)) -> AP(ap(f, x'''), x0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

AP(ap(ap(foldr, ap(f, x''')), h), ap(ap(cons, x0), xs)) -> AP(ap(f, x'''), x0)
AP(ap(ap(foldr, ap(ap(foldr, g''), h'')), h), ap(ap(cons, ap(ap(cons, x''), xs'')), xs)) -> AP(ap(ap(foldr, g''), h''), ap(ap(cons, x''), xs''))
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(ap(g, x), ap(ap(ap(foldr, g), h), xs))
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:
0:01 minutes