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.

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 contains 1 SCC with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly 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(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)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app2(x1, x2)
cons  =  cons
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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.