Termination w.r.t. Q proof of TRS/AG01#3.12.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(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

  ↳ 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)))

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:

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)))

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:

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)))

Q is empty.
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

  ↳ 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)))

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 oriented strictly and are deleted.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP₂(x₁, x₂)) = 3·x₁
POL(add₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ 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)))

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

        ↳ 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)))

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 oriented strictly and are deleted.

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REVERSE₁(x₁)) = 3·x₁
POL(add₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ 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)))

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

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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)))

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SHUFFLE₁(x₁)) = x₁
POL(add₂(x₁, x₂)) = 3 + x₂
POL(app₂(x₁, x₂)) = x₁ + x₂
POL(nil) = 0
POL(reverse₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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)))

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.