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:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(takeWhile, p), xs))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(dropWhile, p), xs))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs)))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(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:

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(takeWhile, p), xs))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(dropWhile, p), xs))
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs)))
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, true), x), y) → y
app(app(takeWhile, p), nil) → nil
app(app(takeWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(takeWhile, p), xs))), nil)
app(app(dropWhile, p), nil) → nil
app(app(dropWhile, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(dropWhile, p), xs)), app(app(cons, x), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QDPSizeChangeProof

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

APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(dropWhile, p), app(app(cons, x), xs)) → APP(app(dropWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(app(takeWhile, p), xs)
APP(app(takeWhile, p), app(app(cons, x), xs)) → APP(p, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: