Termination w.r.t. Q proof of TRS/currying/D3333.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(h, z), app(e, x)) → APP(c, z)
APP(app(g, app(e, x)), app(e, y)) → APP(g, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(d, z), app(app(g, 0), 0)) → APP(e, 0)
APP(app(g, app(e, x)), app(e, y)) → APP(e, app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(h, z), app(e, x)) → APP(h, app(c, z))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(h, z), app(e, x)) → APP(d, z)
APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)
APP(app(d, z), app(app(g, x), y)) → APP(e, 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(d, z), app(app(g, x), y)) → APP(g, app(e, x))
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(h, z), app(e, x)) → APP(app(d, z), x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(d, z), app(app(g, x), y)) → APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(d, z)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(h, z), app(e, x)) → APP(c, z)
APP(app(g, app(e, x)), app(e, y)) → APP(g, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(d, z), app(app(g, 0), 0)) → APP(e, 0)
APP(app(g, app(e, x)), app(e, y)) → APP(e, app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(h, z), app(e, x)) → APP(h, app(c, z))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(h, z), app(e, x)) → APP(d, z)
APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)
APP(app(d, z), app(app(g, x), y)) → APP(e, 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(d, z), app(app(g, x), y)) → APP(g, app(e, x))
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(h, z), app(e, x)) → APP(app(d, z), x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(d, z), app(app(g, x), y)) → APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(d, z)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(g, app(e, x)), app(e, y)) → APP(g, x)
APP(app(h, z), app(e, x)) → APP(c, z)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(g, app(e, x)), app(e, y)) → APP(e, app(app(g, x), y))
APP(app(d, z), app(app(g, 0), 0)) → APP(e, 0)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(h, z), app(e, x)) → APP(h, app(c, z))
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(h, z), app(e, x)) → APP(d, z)
APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)
APP(app(d, z), app(app(g, x), y)) → APP(e, 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(d, z), app(app(g, x), y)) → APP(g, app(e, x))
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(h, z), app(e, x)) → APP(app(d, z), x)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(d, z), app(app(g, x), y)) → APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(d, z)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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 23 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(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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:

G(e(x), e(y)) → G(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(g, app(e, x)), app(e, y)) → APP(app(g, x), y)

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

Lexicographic path order with status [19].
Precedence:

e₁ > G₁

Status:

G₁: [1]
e₁: multiset

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(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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)

D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))

The TRS R consists of the following rules:

g(e(x), e(y)) → e(g(x, y))

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(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))

The remaining pairs can at least be oriented weakly.

APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))

Used ordering: Combined order from the following AFS and order.
D(x1, x2) = D(x1)
g(x1, x2) = g(x1, x2)
c(x1) = c(x1)
0 = 0
e(x1) = e

Lexicographic path order with status [19].
Precedence:

0 > D₁ > g₂
0 > c₁ > g₂
e > g₂

Status:

e: []
g₂: [1,2]
D₁: [1]
c₁: multiset
0: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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)

D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))

The TRS R consists of the following rules:

g(e(x), e(y)) → e(g(x, y))

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(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D(x1, x2) = x2
g(x1, x2) = g(x1, x2)
c(x1) = c
0 = 0
e(x1) = e(x1)

Lexicographic path order with status [19].
Precedence:

g₂ > e₁

Status:

g₂: [2,1]
c: multiset
0: multiset
e₁: [1]

The following usable rules [14] were oriented:

app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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

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

APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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

        ↳ 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(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(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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(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].
Precedence:

APP₁ > map > app₂ > filter
filter2 > filter
true > app₂ > filter
cons > app₂ > filter
false > app₂ > filter

Status:

filter: multiset
true: multiset
APP₁: [1]
map: multiset
false: multiset
app₂: [1,2]
filter2: multiset
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ 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(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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)

MAP(f, cons(x, xs)) → MAP(f, xs)

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(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(x1, x2)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > MAP₂

Status:

cons₂: multiset
MAP₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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.