Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/secret2005SLASHaprove1.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:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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(map, f), xs) -> AP₂(ap₂(if, ap₂(isEmpty, xs)), f)
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(f, ap₂(last, xs))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(ap₂(if2, f), xs)
AP₂(ap₂(map, f), xs) -> AP₂(if, ap₂(isEmpty, xs))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(cons, ap₂(f, ap₂(last, xs)))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
AP₂(ap₂(map, f), xs) -> AP₂(isEmpty, xs)
AP₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(last, ap₂(ap₂(cons, y), ys))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(last, xs)
AP₂(ap₂(if2, f), xs) -> AP₂(dropLast, xs)
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(if2, f)
AP₂(ap₂(map, f), xs) -> AP₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
AP₂(ap₂(if2, f), xs) -> AP₂(ap₂(map, f), ap₂(dropLast, xs))
AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(dropLast, ap₂(ap₂(cons, y), ys))
AP₂(ap₂(if2, f), xs) -> AP₂(map, f)
AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The TRS R consists of the following rules:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

AP₂(ap₂(map, f), xs) -> AP₂(ap₂(if, ap₂(isEmpty, xs)), f)
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(f, ap₂(last, xs))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(ap₂(if2, f), xs)
AP₂(ap₂(map, f), xs) -> AP₂(if, ap₂(isEmpty, xs))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(cons, ap₂(f, ap₂(last, xs)))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
AP₂(ap₂(map, f), xs) -> AP₂(isEmpty, xs)
AP₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(last, ap₂(ap₂(cons, y), ys))
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(last, xs)
AP₂(ap₂(if2, f), xs) -> AP₂(dropLast, xs)
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(if2, f)
AP₂(ap₂(map, f), xs) -> AP₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
AP₂(ap₂(if2, f), xs) -> AP₂(ap₂(map, f), ap₂(dropLast, xs))
AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(dropLast, ap₂(ap₂(cons, y), ys))
AP₂(ap₂(if2, f), xs) -> AP₂(map, f)
AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The TRS R consists of the following rules:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 10 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(dropLast, ap₂(ap₂(cons, y), ys))

The TRS R consists of the following rules:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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

The following pairs can be strictly oriented and are deleted.

AP₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(dropLast, ap₂(ap₂(cons, y), ys))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
AP₂(x1, x2) = AP₁(x2)
dropLast = dropLast
ap₂(x1, x2) = ap₂(x1, x2)
cons = cons

Lexicographic Path Order [19].
Precedence:

dropLast > [AP₁, ap_2] > cons

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

AP₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(last, ap₂(ap₂(cons, y), ys))

The TRS R consists of the following rules:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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

The following pairs can be strictly oriented and are deleted.

AP₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> AP₂(last, ap₂(ap₂(cons, y), ys))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
AP₂(x1, x2) = AP₁(x2)
last = last
ap₂(x1, x2) = ap₂(x1, x2)
cons = cons

Lexicographic Path Order [19].
Precedence:

last > [AP₁, ap_2] > cons

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(f, ap₂(last, xs))
AP₂(ap₂(map, f), xs) -> AP₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
AP₂(ap₂(ap₂(if, null), f), xs) -> AP₂(ap₂(if2, f), xs)
AP₂(ap₂(if2, f), xs) -> AP₂(ap₂(map, f), ap₂(dropLast, xs))

The TRS R consists of the following rules:

ap₂(ap₂(map, f), xs) -> ap₂(ap₂(ap₂(if, ap₂(isEmpty, xs)), f), xs)
ap₂(ap₂(ap₂(if, true), f), xs) -> null
ap₂(ap₂(ap₂(if, null), f), xs) -> ap₂(ap₂(cons, ap₂(f, ap₂(last, xs))), ap₂(ap₂(if2, f), xs))
ap₂(ap₂(if2, f), xs) -> ap₂(ap₂(map, f), ap₂(dropLast, xs))
ap₂(isEmpty, null) -> true
ap₂(isEmpty, ap₂(ap₂(cons, x), xs)) -> null
ap₂(last, ap₂(ap₂(cons, x), null)) -> x
ap₂(last, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(last, ap₂(ap₂(cons, y), ys))
ap₂(dropLast, ap₂(ap₂(cons, x), null)) -> null
ap₂(dropLast, ap₂(ap₂(cons, x), ap₂(ap₂(cons, y), ys))) -> ap₂(ap₂(cons, x), ap₂(dropLast, ap₂(ap₂(cons, y), ys)))

The set Q consists of the following terms:

ap₂(ap₂(map, x0), x1)
ap₂(ap₂(ap₂(if, true), x0), x1)
ap₂(ap₂(ap₂(if, null), x0), x1)
ap₂(ap₂(if2, x0), x1)
ap₂(isEmpty, null)
ap₂(isEmpty, ap₂(ap₂(cons, x0), x1))
ap₂(last, ap₂(ap₂(cons, x0), null))
ap₂(last, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))
ap₂(dropLast, ap₂(ap₂(cons, x0), null))
ap₂(dropLast, ap₂(ap₂(cons, x0), ap₂(ap₂(cons, x1), x2)))

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