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, ap2] > 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, ap2] > 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.