Termination w.r.t. Q proof of TRS/higher-order/LifantsevEx3Lists.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(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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 7 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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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:

APPEND(cons(h, t), l) → APPEND(t, l)
APPEND(append(l1, l2), l3) → APPEND(l2, l3)
APPEND(append(l1, l2), l3) → APPEND(l1, append(l2, l3))

The TRS R consists of the following rules:

append(nil, l) → l
append(cons(h, t), l) → cons(h, append(t, l))
append(append(l1, l2), l3) → append(l1, append(l2, l3))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))

The remaining pairs can at least be oriented weakly.

APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2) = x1
cons(x1, x2) = x2
append(x1, x2) = append(x1, x2)
nil = nil

Lexicographic Path Order [19].
Precedence:

nil > append₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

APPEND(cons(h, t), l) → APPEND(t, l)

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(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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, h), t)) → APP(f, h)

The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = x1
app(x1, x2) = app(x1, x2)
map = map
append = append
cons = cons

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

                    ↳ QDPOrderProof

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

APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

MAP(f, cons(h, t)) → MAP(f, t)
MAP(f, append(l1, l2)) → MAP(f, l2)
MAP(f, append(l1, l2)) → MAP(f, l1)

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(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

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)
append(x1, x2) = append(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > MAP₂
append₂ > MAP₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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.