Termination w.r.t. Q proof of TRS/currying/D3308.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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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

  ↳ Overlay + Local Confluence

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
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(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
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(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
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(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, 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(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APP(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
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(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
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(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
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(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, 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(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

APP(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
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(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
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(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
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(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
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(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 19 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(D, y)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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:

D1(+(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
D1(+(x, y)) → D1(y)
D1(-(x, y)) → D1(x)

R is empty.
The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(D, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D1(x1) = D1(x1)
+(x1, x2) = +(x1, x2)
*(x1, x2) = *(x1, x2)
-(x1, x2) = -(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

+₂: [1,2]
-₂: [2,1]
D1₁: [1]
*₂: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ 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(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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = APP(x1)
app(x1, x2) = app(x1, x2)
filter2 = filter2
true = true
filter = filter
map = map
cons = cons
false = false

Lexicographic path order with status [19].
Quasi-Precedence:

true > filter > [APP₁, app₂, filter2, cons]
map > [APP₁, app₂, filter2, cons]
false > filter > [APP₁, app₂, filter2, cons]

Status:

filter: multiset
true: multiset
APP₁: [1]
map: multiset
false: multiset
app₂: [2,1]
filter2: multiset
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

MAP(f, cons(x, xs)) → MAP(f, xs)

R is empty.
The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

cons₂ > MAP₁

Status:

MAP₁: [1]
cons₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.