Termination w.r.t. Q proof of TRS/higher-order/AotoYam026.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(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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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(twice, f) → APP(comp, f)
APP(twice, f) → APP(app(comp, f), f)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(comp, f), g), x) → APP(g, x)

The TRS R consists of the following rules:

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APP(twice, f) → APP(comp, f)
APP(twice, f) → APP(app(comp, f), f)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(comp, f), g), x) → APP(g, x)

The TRS R consists of the following rules:

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(twice, f) → APP(app(comp, f), f)
APP(twice, f) → APP(comp, f)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(comp, f), g), x) → APP(g, x)

The TRS R consists of the following rules:

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(comp, f), g), x) → APP(g, x)

The TRS R consists of the following rules:

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(app(comp, f), g), x) → APP(g, 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)
map = map
cons = cons
comp = comp
nil = nil
twice = twice

Lexicographic path order with status [19].
Quasi-Precedence:

map > [APP₁, cons, comp] > [app₂, twice] > nil

Status:

comp: multiset
twice: multiset
APP₁: [1]
map: multiset
app₂: [1,2]
nil: multiset
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ 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(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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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:

MAP(f, cons(x, xs)) → MAP(f, xs)

R is empty.
The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
comp(x0, x1, x2)
twice(x0)

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

Lexicographic path order with status [19].
Quasi-Precedence:

cons₂ > MAP₁

Status:

MAP₁: [1]
cons₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

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(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.