Termination w.r.t. Q proof of /tmp/tmp9d4UIA/nonTermF.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following 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))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following 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))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

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)
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))

The TRS R consists of the following 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.