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:

ap₂(ap₂(f, x), x) -> ap₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
ap₂(ap₂(ap₂(foldr, g), h), nil) -> h
ap₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> ap₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ap₂(ap₂(f, x), x) -> ap₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
ap₂(ap₂(ap₂(foldr, g), h), nil) -> h
ap₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> ap₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AP₂(ap₂(f, x), x) -> AP₂(x, ap₂(f, x))
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(g, x)
AP₂(ap₂(f, x), x) -> AP₂(ap₂(cons, x), nil)
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))
AP₂(ap₂(f, x), x) -> AP₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
AP₂(ap₂(f, x), x) -> AP₂(cons, x)
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(ap₂(foldr, g), h), xs)

The TRS R consists of the following rules:

ap₂(ap₂(f, x), x) -> ap₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
ap₂(ap₂(ap₂(foldr, g), h), nil) -> h
ap₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> ap₂(ap₂(g, x), ap₂(ap₂(ap₂(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:

AP₂(ap₂(f, x), x) -> AP₂(x, ap₂(f, x))
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(g, x)
AP₂(ap₂(f, x), x) -> AP₂(ap₂(cons, x), nil)
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))
AP₂(ap₂(f, x), x) -> AP₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
AP₂(ap₂(f, x), x) -> AP₂(cons, x)
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(ap₂(foldr, g), h), xs)

The TRS R consists of the following rules:

ap₂(ap₂(f, x), x) -> ap₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
ap₂(ap₂(ap₂(foldr, g), h), nil) -> h
ap₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> ap₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

AP₂(ap₂(f, x), x) -> AP₂(x, ap₂(f, x))
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(g, x)
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))
AP₂(ap₂(f, x), x) -> AP₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
AP₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> AP₂(ap₂(ap₂(foldr, g), h), xs)

The TRS R consists of the following rules:

ap₂(ap₂(f, x), x) -> ap₂(ap₂(x, ap₂(f, x)), ap₂(ap₂(cons, x), nil))
ap₂(ap₂(ap₂(foldr, g), h), nil) -> h
ap₂(ap₂(ap₂(foldr, g), h), ap₂(ap₂(cons, x), xs)) -> ap₂(ap₂(g, x), ap₂(ap₂(ap₂(foldr, g), h), xs))

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