Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/higher-orderSLASHAotoYamSLASH012.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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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

  ↳ DependencyPairsProof

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

app₂(app₂(and, true), true) -> true
app₂(app₂(and, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.

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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
We have to consider all minimal (P,Q,R)-chains.

↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
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

  ↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
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

  ↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
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

  ↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
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

      ↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
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

  ↳ 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, x), false) -> false
app₂(app₂(and, false), y) -> false
app₂(app₂(or, true), y) -> true
app₂(app₂(or, x), 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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.