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

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), 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:

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(dropWhile, p), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs)))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(takeWhile, p), xs))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(dropWhile, p), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs)))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(takeWhile, p), xs))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(dropWhile, p), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs)))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(takeWhile, p), xs))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), 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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)

The remaining pairs can at least be oriented weakly.

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = APP(x1)
app(x1, x2) = app(x1, x2)
dropWhile = dropWhile
cons = cons
takeWhile = takeWhile

Lexicographic path order with status [19].
Precedence:

cons > dropWhile > APP₁
takeWhile > app₂ > dropWhile > APP₁

Status:

APP₁: [1]
app₂: [1,2]
dropWhile: multiset
takeWhile: multiset
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                      ↳ QDP

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

APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

TAKEWHILE(p, cons(x, xs)) → TAKEWHILE(p, xs)

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

The following pairs can be oriented strictly and are deleted.

APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TAKEWHILE(x1, x2) = TAKEWHILE(x1, x2)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > TAKEWHILE₂

Status:

cons₂: multiset
TAKEWHILE₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                      ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

DROPWHILE(p, cons(x, xs)) → DROPWHILE(p, xs)

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

The following pairs can be oriented strictly and are deleted.

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DROPWHILE(x1, x2) = DROPWHILE(x1, x2)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > DROPWHILE₂

Status:

DROPWHILE₂: [1,2]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.