Termination w.r.t. Q proof of TRS/LJB01jones5.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:

f₂(x, empty) -> x
f₂(empty, cons₂(a, k)) -> f₂(cons₂(a, k), k)
f₂(cons₂(a, k), y) -> f₂(y, k)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, empty) -> x
f₂(empty, cons₂(a, k)) -> f₂(cons₂(a, k), k)
f₂(cons₂(a, k), y) -> f₂(y, k)

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:

F₂(cons₂(a, k), y) -> F₂(y, k)
F₂(empty, cons₂(a, k)) -> F₂(cons₂(a, k), k)

The TRS R consists of the following rules:

f₂(x, empty) -> x
f₂(empty, cons₂(a, k)) -> f₂(cons₂(a, k), k)
f₂(cons₂(a, k), y) -> f₂(y, k)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

F₂(cons₂(a, k), y) -> F₂(y, k)
F₂(empty, cons₂(a, k)) -> F₂(cons₂(a, k), k)

The TRS R consists of the following rules:

f₂(x, empty) -> x
f₂(empty, cons₂(a, k)) -> f₂(cons₂(a, k), k)
f₂(cons₂(a, k), y) -> f₂(y, k)

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