Termination w.r.t. Q proof of /tmp/tmpKsnWuj/31.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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(:, app(app(+, x), y)), z) → APP(:, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
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(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
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(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
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(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
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(:, z), app(app(+, x), app(f, y))) → APP(g, z)
APP(app(:, app(app(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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

      ↳ DependencyGraphProof

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

APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(:, app(app(+, x), y)), z) → APP(:, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
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(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
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(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
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(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
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(:, z), app(app(+, x), app(f, y))) → APP(g, z)
APP(app(:, app(app(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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 [15,17,22] contains 2 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)

The TRS R consists of the following rules:

app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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 [15].

The following pairs can be oriented strictly and are deleted.

APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(+) = 0
POL(:) = 0
POL(APP(x₁, x₂)) = x₁
POL(a) = 0
POL(app(x₁, x₂)) = 1 + x₁ + x₂
POL(g) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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

      ↳ 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(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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