Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/higher-orderSLASHAotoYamSLASH009.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:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(and, app₂(p, x))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(or, app₂(p, x))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(and, app₂(p, x))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(or, app₂(p, x))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(p, x)

The remaining pairs can at least by weakly be oriented.

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = x1
app₂(x1, x2) = app₁(x2)
forall = forall
cons = cons
forsome = forsome

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)
APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                      ↳ QDP

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

APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

APP₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forsome, p), xs)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x2)
app₂(x1, x2) = app₁(x2)
forsome = forsome
cons = cons

Lexicographic Path Order [19].
Precedence:

[APP₁, app_1] > [forsome, cons]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                      ↳ QDP

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

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

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

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)

The TRS R consists of the following rules:

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

APP₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> APP₂(app₂(forall, p), xs)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x2)
app₂(x1, x2) = app₁(x2)
forall = forall
cons = cons

Lexicographic Path Order [19].
Precedence:

[APP₁, app_1] > [forall, cons]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

app₂(app₂(and, true), true) -> true
app₂(app₂(and, true), false) -> false
app₂(app₂(and, false), true) -> false
app₂(app₂(and, false), false) -> false
app₂(app₂(or, true), true) -> true
app₂(app₂(or, true), false) -> true
app₂(app₂(or, false), true) -> true
app₂(app₂(or, false), false) -> false
app₂(app₂(forall, p), nil) -> true
app₂(app₂(forall, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(and, app₂(p, x)), app₂(app₂(forall, p), xs))
app₂(app₂(forsome, p), nil) -> false
app₂(app₂(forsome, p), app₂(app₂(cons, x), xs)) -> app₂(app₂(or, app₂(p, x)), app₂(app₂(forsome, p), xs))

The set Q consists of the following terms:

app₂(app₂(and, true), true)
app₂(app₂(and, true), false)
app₂(app₂(and, false), true)
app₂(app₂(and, false), false)
app₂(app₂(or, true), true)
app₂(app₂(or, true), false)
app₂(app₂(or, false), true)
app₂(app₂(or, false), false)
app₂(app₂(forall, x0), nil)
app₂(app₂(forall, x0), app₂(app₂(cons, x1), x2))
app₂(app₂(forsome, x0), nil)
app₂(app₂(forsome, x0), app₂(app₂(cons, x1), x2))

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