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
  ↳ 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)))

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(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)))

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(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)))

Q is empty.
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
  ↳ 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)))

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.


AP2(dropLast, ap2(ap2(cons, x), ap2(ap2(cons, y), ys))) -> AP2(dropLast, ap2(ap2(cons, y), ys))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(AP2(x1, x2)) = 2·x2   
POL(ap2(x1, x2)) = 2 + x1 + 2·x2   
POL(cons) = 0   
POL(dropLast) = 0   

The following usable rules [14] were oriented: none



↳ 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)))

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
      ↳ 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)))

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.


AP2(last, ap2(ap2(cons, x), ap2(ap2(cons, y), ys))) -> AP2(last, ap2(ap2(cons, y), ys))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(AP2(x1, x2)) = 2·x2   
POL(ap2(x1, x2)) = 2 + x1 + 2·x2   
POL(cons) = 0   
POL(last) = 0   

The following usable rules [14] were oriented: none



↳ 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)))

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

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)))

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.


AP2(ap2(ap2(if, null), f), xs) -> AP2(f, ap2(last, xs))
The remaining pairs can at least be oriented weakly.

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))
Used ordering: Polynomial interpretation [21]:

POL(AP2(x1, x2)) = 2·x1   
POL(ap2(x1, x2)) = 2 + 2·x2   
POL(cons) = 0   
POL(dropLast) = 2   
POL(if) = 2   
POL(if2) = 0   
POL(isEmpty) = 1   
POL(last) = 0   
POL(map) = 2   
POL(null) = 1   
POL(true) = 2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP

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

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)))

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