Termination w.r.t. Q proof of TRS/currying/D3312.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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(and, app(app(or, y), z)), x) → APP(app(and, x), z)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(and, x), app(app(or, y), z)) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(not, app(app(or, x), y)) → APP(app(and, app(not, x)), app(not, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(and, x), app(app(or, y), z)) → APP(or, app(app(and, x), y))
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(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
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(and, app(app(or, y), z)), x) → APP(and, x)
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(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(or, app(not, x))
APP(not, app(app(or, x), y)) → APP(and, app(not, x))
APP(app(and, app(app(or, y), z)), x) → APP(or, app(app(and, x), y))
APP(app(and, app(app(or, y), z)), x) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(not, app(app(and, x), y)) → APP(app(or, app(not, x)), app(not, y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(and, app(app(or, y), z)), x) → APP(app(and, x), z)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(and, x), app(app(or, y), z)) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(not, app(app(or, x), y)) → APP(app(and, app(not, x)), app(not, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(and, x), app(app(or, y), z)) → APP(or, app(app(and, x), y))
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(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
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(and, app(app(or, y), z)), x) → APP(and, x)
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(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(or, app(not, x))
APP(not, app(app(or, x), y)) → APP(and, app(not, x))
APP(app(and, app(app(or, y), z)), x) → APP(or, app(app(and, x), y))
APP(app(and, app(app(or, y), z)), x) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(not, app(app(and, x), y)) → APP(app(or, app(not, x)), app(not, y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(and, app(app(or, y), z)), x) → APP(app(and, x), z)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
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(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(not, app(app(or, x), y)) → APP(app(and, app(not, x)), app(not, y))
APP(app(and, x), app(app(or, y), z)) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(and, x), app(app(or, y), z)) → APP(or, app(app(and, x), y))
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(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), 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(and, app(app(or, y), z)), x) → APP(and, x)
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(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(not, app(app(or, x), y)) → APP(not, x)
APP(app(and, app(app(or, y), z)), x) → APP(or, app(app(and, x), y))
APP(not, app(app(or, x), y)) → APP(and, app(not, x))
APP(not, app(app(and, x), y)) → APP(or, app(not, x))
APP(app(and, app(app(or, y), z)), x) → APP(app(or, app(app(and, x), y)), app(app(and, x), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(not, app(app(and, x), y)) → APP(app(or, app(not, x)), app(not, y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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 18 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(and, app(app(or, y), z)), x) → APP(app(and, x), z)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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:

AND(x, or(y, z)) → AND(x, z)
AND(x, or(y, z)) → AND(x, y)
AND(or(y, z), x) → AND(x, y)
AND(or(y, z), x) → AND(x, z)

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(and, app(app(or, y), z)), x) → APP(app(and, x), z)
APP(app(and, app(app(or, y), z)), x) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), y)
APP(app(and, x), app(app(or, y), z)) → APP(app(and, x), z)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

AND₂: multiset

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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)

The TRS R consists of the following rules:

app(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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)

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(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(not, app(app(or, x), y)) → APP(not, x)
APP(not, app(app(or, x), y)) → APP(not, y)
APP(not, app(app(and, x), y)) → APP(not, x)
APP(not, app(app(and, x), y)) → APP(not, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NOT(x1) = x1
and(x1, x2) = and(x1, x2)
or(x1, x2) = or(x1, x2)

Recursive path order with status [2].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ 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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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

Recursive path order with status [2].
Precedence:

trivial

Status:

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(not, app(not, x)) → x
app(not, app(app(or, x), y)) → app(app(and, app(not, x)), app(not, y))
app(not, app(app(and, x), y)) → app(app(or, app(not, x)), app(not, y))
app(app(and, x), app(app(or, y), z)) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
app(app(and, app(app(or, y), z)), x) → app(app(or, app(app(and, x), y)), app(app(and, x), z))
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.