Termination w.r.t. Q proof of TRS/currying/AG01#3.22.trs

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(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(times, app(s, z))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, app(s, z)), 0)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(times, app(s, z))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, app(s, z)), 0)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(times, app(s, z))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, app(s, z)), 0)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

PLUS(x, s(y)) → PLUS(x, y)

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

The following pairs can be oriented strictly and are deleted.

APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = x2
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, s(y)) → TIMES(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

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

The following pairs can be oriented strictly and are deleted.

APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(times, x), app(app(plus, y), app(s, z))) → APP(app(times, x), app(s, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2) = x2
plus(x1, x2) = plus(x1, x2)
s(x1) = s(x1)
times(x1, x2) = times
0 = 0

Lexicographic Path Order [19].
Precedence:

plus₂ > s₁ > times > 0

The following usable rules [14] were oriented:

app(app(times, x), 0) → 0
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, x), 0) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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.

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The remaining pairs can at least be oriented weakly.

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = x2
app(x1, x2) = app(x1, x2)
cons = cons

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(times, x), app(app(plus, y), app(s, z))) → app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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 2 less nodes.