Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/AG01SLASHHASH3.12.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

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(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, 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:

REVERSE₁(add₂(n, x)) -> APP₂(reverse₁(x), add₂(n, nil))
REVERSE₁(add₂(n, x)) -> REVERSE₁(x)
SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))
SHUFFLE₁(add₂(n, x)) -> REVERSE₁(x)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))

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:

REVERSE₁(add₂(n, x)) -> APP₂(reverse₁(x), add₂(n, nil))
REVERSE₁(add₂(n, x)) -> REVERSE₁(x)
SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))
SHUFFLE₁(add₂(n, x)) -> REVERSE₁(x)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))

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(add₂(n, x), y) -> APP₂(x, y)

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(x1)
add₂(x1, x2) = add₁(x2)

Lexicographic Path Order [19].
Precedence:

add₁ > APP₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, 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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The TRS R consists of the following rules:

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))

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.

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REVERSE₁(x1) = REVERSE₁(x1)
add₂(x1, x2) = add₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

add₂ > REVERSE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, 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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))

The TRS R consists of the following rules:

app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))

The set Q consists of the following terms:

app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))

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