Termination w.r.t. Q proof of /tmp/tmphuCeLv/#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:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(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:

SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (4)x_1
POL(add(x₁, x₂)) = 2 + (4)x_2

The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(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:

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

REVERSE(add(n, x)) → REVERSE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 2 + (4)x_2
POL(REVERSE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(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:

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SHUFFLE(x₁)) = (1/4)x_1
POL(add(x₁, x₂)) = 1/4 + (1/2)x_1 + x_2
POL(reverse(x₁)) = x_1
POL(app(x₁, x₂)) = x_1 + x_2
POL(nil) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(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:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(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.