Termination w.r.t. Q proof of TRS/currying/D3330.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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), 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(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(app(app(filter2, app(f, x)), f), x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
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)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), 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(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(app(app(filter2, app(f, x)), f), x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
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)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
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(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
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(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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, x)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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 14 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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)

The TRS R consists of the following rules:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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:

:1(:(:(:(C, x), y), z), u) → :1(:(x, z), :(:(:(x, y), z), u))
:1(:(:(:(C, x), y), z), u) → :1(x, z)
:1(:(:(:(C, x), y), z), u) → :1(:(x, y), z)
:1(:(:(:(C, x), y), z), u) → :1(x, y)
:1(:(:(:(C, x), y), z), u) → :1(:(:(x, y), z), u)

The TRS R consists of the following rules:

:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))

The set Q consists of the following terms:

:(:(:(:(C, x0), x1), x2), x3)
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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), y)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, x), y)), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, x), z)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(app(:, app(app(:, app(app(:, x), y)), z)), u)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
:1(x1, x2) = :1(x1, x2)
:(x1, x2) = :(x1, x2)
C = C

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

[:1₂, :₂, C]

Status:

:1₂: [1,2]
:₂: [1,2]
C: multiset

The following usable rules [14] were oriented:

app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))

↳ 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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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:

:(:(:(:(C, x0), x1), x2), x3)
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(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u))
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(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3)
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.