Termination w.r.t. Q proof of /tmp/tmp

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 [19] 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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
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), y)
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(:, 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)
APP(app(map, 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(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
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(app(:, x), z)), 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(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), y)), z)
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(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), 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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
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), y)
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(:, 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)
APP(app(map, 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(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, app(app(:, x), y)), z))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) → APP(:, app(app(:, x), z))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
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(app(:, x), z)), 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(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), y)), z)
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(:, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), 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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 14 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

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

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

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

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))

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 have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

              ↳ QDP

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

:1(:(:(:(C, x), y), z), u) → :1(x, z)
:1(:(:(:(C, x), y), z), u) → :1(x, y)
:1(:(:(:(C, x), y), z), u) → :1(:(x, y), z)
:1(:(:(:(C, x), y), z), u) → :1(:(:(x, y), z), u)
:1(:(:(:(C, x), y), z), u) → :1(:(x, z), :(:(:(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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPOrderProof

              ↳ QDP

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

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

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

Recursive path order with status [2].
Quasi-Precedence:

:1₂ > [:₂, C]

Status:

:₂: [1,2]
:1₂: [1,2]
C: multiset

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
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)

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

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

APP(app(filter, 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(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(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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

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

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

R is empty.
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.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2