Termination w.r.t. Q proof of TRS/higher-order/AProVE

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

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

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(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(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
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(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(minus, app(s, x)), app(s, y)) → APP(minus, x)
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(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

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 with status [19].
Precedence:

s₁ > MINUS₁

Status:

MINUS₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

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

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

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

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Precedence:

s₁ > DIV₁

Status:

DIV₁: [1]
s₁: [1]
0: multiset

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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)

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

Lexicographic path order with status [19].
Precedence:

app₂ > APP₁ > map
cons > APP₁ > map

Status:

APP₁: [1]
map: multiset
app₂: [1,2]
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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