Termination w.r.t. Q proof of TRS/currying/AG01#3.18.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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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

  ↳ DependencyPairsProof

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.

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(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(double, app(s, x)) → APP(s, app(double, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(plus, app(s, x)), y) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
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(plus, app(s, x)), y) → APP(plus, x)
APP(double, app(s, x)) → APP(s, app(s, app(double, x)))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(double, app(s, x)) → APP(double, 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(app(plus, app(s, x)), y) → APP(app(minus, x), 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(app(plus, app(s, x)), y) → APP(double, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(s, y)
APP(app(plus, app(s, x)), y) → APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
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(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(double, app(s, x)) → APP(s, app(double, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(plus, app(s, x)), y) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
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(plus, app(s, x)), y) → APP(plus, x)
APP(double, app(s, x)) → APP(s, app(s, app(double, x)))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(double, app(s, x)) → APP(double, 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(app(plus, app(s, x)), y) → APP(app(minus, x), 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(app(plus, app(s, x)), y) → APP(double, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(s, y)
APP(app(plus, app(s, x)), y) → APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
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(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
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(plus, app(s, x)), y) → APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(double, app(s, x)) → APP(s, app(double, x))
APP(app(plus, app(s, x)), y) → APP(minus, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))
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(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(double, app(s, x)) → APP(s, app(s, app(double, x)))
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(double, app(s, x)) → APP(double, 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(app(plus, app(s, x)), y) → APP(app(minus, x), 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(app(plus, app(s, x)), y) → APP(double, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(s, x)), y) → APP(s, y)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, app(app(minus, x), y))
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(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

APP(double, app(s, x)) → APP(double, x)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
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:

DOUBLE(S(x)) → DOUBLE(x)

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

The following pairs can be oriented strictly and are deleted.

APP(double, app(s, x)) → APP(double, x)

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

Lexicographic Path Order [19].
Precedence:

[DOUBLE₁, S_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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)

MINUS(s(x), s(y)) → MINUS(x, y)

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

The following pairs can be oriented strictly and are deleted.

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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)

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))

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

The following pairs can be oriented strictly and are deleted.

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(app(plus, x), app(s, y))

The remaining pairs can at least be oriented weakly.

APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))

Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = x1
s(x1) = s(x1)
minus(x1, x2) = minus(x1)
double(x1) = double(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

[double₁, 0] > [s₁, minus_1]

The following usable rules [14] were oriented:

app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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)

PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))

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

The following pairs can be oriented strictly and are deleted.

APP(app(plus, app(s, x)), y) → APP(app(plus, app(app(minus, x), y)), app(double, y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = x1
s(x1) = s(x1)
minus(x1, x2) = x1
double(x1) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ 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(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
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(map, f), app(app(cons, x), xs)) → APP(f, x)
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)

The remaining pairs can at least be oriented weakly.

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)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = APP(x2)
app(x1, x2) = app(x1, x2)
filter2 = filter2
true = true
filter = filter
map = map
cons = cons
false = false
nil = nil
double = double
s = s
plus = plus
0 = 0
minus = minus

Lexicographic Path Order [19].
Precedence:

true > app₂ > filter2
true > app₂ > plus
true > app₂ > 0
map > APP₁ > filter > app₂ > filter2
map > APP₁ > filter > app₂ > plus
map > APP₁ > filter > app₂ > 0
map > cons > app₂ > filter2
map > cons > app₂ > plus
map > cons > app₂ > 0
[double, s] > app₂ > filter2
[double, s] > app₂ > plus
[double, s] > app₂ > 0
minus > app₂ > filter2
minus > app₂ > plus
minus > app₂ > 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(double, 0) → 0
app(double, app(s, x)) → app(s, app(s, app(double, x)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, app(app(minus, x), y)), app(double, 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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.