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:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(cons, x), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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:

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs)))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(cons, x), xs))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(takeWhile, p), xs))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(cons, x), 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:

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs)))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(cons, x), xs))
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(takeWhile, p), xs))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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 strictly oriented and are deleted.


APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(p, x)
The remaining pairs can at least by weakly be oriented.

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app1(x2)
dropWhile  =  dropWhile
cons  =  cons
takeWhile  =  takeWhile

Lexicographic Path Order [19].
Precedence:
[cons, takeWhile] > app1


The following usable rules [14] were oriented: none



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

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

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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 strictly oriented and are deleted.


APP2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(takeWhile, p), xs)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
takeWhile  =  takeWhile
cons  =  cons

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > [takeWhile, cons]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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 strictly oriented and are deleted.


APP2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> APP2(app2(dropWhile, p), xs)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
dropWhile  =  dropWhile
cons  =  cons

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > [dropWhile, cons]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, true), x), y) -> y
app2(app2(takeWhile, p), nil) -> nil
app2(app2(takeWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(takeWhile, p), xs))), nil)
app2(app2(dropWhile, p), nil) -> nil
app2(app2(dropWhile, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(dropWhile, p), xs)), app2(app2(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.