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

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(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(copy, app(s, x)), y), z) → APP(copy, x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), y), z)
APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(copy, app(s, x)), y), z) → APP(cons, app(f, y))
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(copy, app(s, x)), y), z) → APP(copy, x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), y), z)
APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(copy, app(s, x)), y), z) → APP(cons, app(f, y))
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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 18 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPApplicativeOrderProof

          ↳ QDP

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

APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

copy1(s(x), y, z) → copy1(x, y, cons(f(y), z))

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(cons(x₁, x₂)) = x₁
POL(copy(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(copy1(x₁, x₂, x₃)) = x₁
POL(f(x₁)) = 0
POL(n) = 0
POL(nil) = 0
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPApplicativeOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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

            ↳ QDPSizeChangeProof

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

APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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