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:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev2, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev1, x), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(cons, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(rev1, y)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev1, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev2, y)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(cons, x), app2(app2(rev2, y), l))
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(rev, app2(app2(cons, x), l)) -> APP2(cons, app2(app2(rev1, x), l))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev2, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev1, x), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(cons, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(rev1, y)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev1, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev2, y)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(cons, x), app2(app2(rev2, y), l))
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(rev, app2(app2(cons, x), l)) -> APP2(cons, app2(app2(rev1, x), l))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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 19 less nodes.

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

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

APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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 oriented strictly and are deleted.


APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 2}


POL( app2(x1, x2) ) = x1 + x2 + 1


POL( cons ) = 1



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:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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

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

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

APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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 oriented strictly and are deleted.


APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
The remaining pairs can at least be oriented weakly.

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 2}


POL( app2(x1, x2) ) = x1 + x2


POL( filter2 ) = 1


POL( true ) = 2


POL( filter ) = 0


POL( map ) = 0


POL( false ) = 1



The following usable rules [14] were oriented: none



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

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

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), 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 oriented strictly and are deleted.


APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.

APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 2}


POL( app2(x1, x2) ) = x1 + x2 + 1


POL( cons ) = 1



The following usable rules [14] were oriented: none



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

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

APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.