Termination w.r.t. Q proof of TRS/currying/Ste92hydra.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(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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

      ↳ EdgeDeletionProof

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

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

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

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 [13,14,18] contains 2 SCCs with 18 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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 [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:

COPY(s(x), y, z) → COPY(x, y, cons(F(y), z))

The TRS R consists of the following rules:

F(cons(nil, y)) → y
F(cons(F(cons(nil, y)), z)) → copy(n, y, z)

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(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: Combined order from the following AFS and order.
COPY(x1, x2, x3) = COPY(x1, x3)
s(x1) = s(x1)
cons(x1, x2) = cons
F(x1) = F(x1)
nil = nil
copy(x1, x2, x3) = copy(x1, x2, x3)
n = n

Lexicographic path order with status [19].
Precedence:

s₁ > COPY₂ > F₁
s₁ > cons > F₁
nil > copy₃ > F₁
n > F₁

Status:

COPY₂: [2,1]
copy₃: multiset
s₁: multiset
nil: multiset
F₁: multiset
n: multiset
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
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, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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)

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
copy = copy
0 = 0
f = f
n = n
s = s

Lexicographic path order with status [19].
Precedence:

APP₁ > app₂
filter2 > app₂
true > app₂
filter > app₂
map > app₂
cons > app₂
false > app₂
nil > app₂
copy > app₂
0 > app₂
f > app₂
s > app₂

Status:

APP₁: [1]
map: multiset
0: multiset
s: multiset
nil: multiset
cons: multiset
filter: multiset
true: multiset
false: multiset
app₂: [2,1]
f: multiset
filter2: multiset
copy: multiset
n: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

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)

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 [13,14,18] contains 0 SCCs with 2 less nodes.