Termination w.r.t. Q proof of TRS/higher-order/AProVE

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

The set Q consists of the following terms:

ap(ap(f, x0), x0)
ap(ap(ap(foldr, x0), x1), nil)
ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3))

Using Dependency Pairs [1,13] 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(x, ap(f, x))
AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) → AP(g, x)
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(f, x), x) → AP(ap(x, ap(f, 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(ap(foldr, g), h), xs)

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))

The set Q consists of the following terms:

ap(ap(f, x0), x0)
ap(ap(ap(foldr, x0), x1), nil)
ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

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

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))

The set Q consists of the following terms:

ap(ap(f, x0), x0)
ap(ap(ap(foldr, x0), x1), nil)
ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

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))

The set Q consists of the following terms:

ap(ap(f, x0), x0)
ap(ap(ap(foldr, x0), x1), nil)
ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(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(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)

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))

The set Q consists of the following terms:

ap(ap(f, x0), x0)
ap(ap(ap(foldr, x0), x1), nil)
ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3))

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