Termination w.r.t. Q proof of TRS/LJB01jones6.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:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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:

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))
F₂(a, empty) -> G₂(a, empty)
F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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:

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))
F₂(a, empty) -> G₂(a, empty)
F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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 with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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.

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))

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

POL(G₂(x₁, x₂)) = 2·x₁
POL(cons₂(x₁, x₂)) = 2 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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

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

F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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.

F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

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

POL(F₂(x₁, x₂)) = 2·x₂
POL(cons₂(x₁, x₂)) = 2 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

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.