Termination w.r.t. Q proof of TRS/Zantema06test1.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(s₁(x), y) -> f₂(x, s₁(s₁(x)))
f₂(x, s₁(s₁(y))) -> f₂(y, x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(s₁(x), y) -> f₂(x, s₁(s₁(x)))
f₂(x, s₁(s₁(y))) -> f₂(y, 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:

F₂(s₁(x), y) -> F₂(x, s₁(s₁(x)))
F₂(x, s₁(s₁(y))) -> F₂(y, x)

The TRS R consists of the following rules:

f₂(s₁(x), y) -> f₂(x, s₁(s₁(x)))
f₂(x, s₁(s₁(y))) -> f₂(y, x)

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(s₁(x), y) -> F₂(x, s₁(s₁(x)))
F₂(x, s₁(s₁(y))) -> F₂(y, x)

The TRS R consists of the following rules:

f₂(s₁(x), y) -> f₂(x, s₁(s₁(x)))
f₂(x, s₁(s₁(y))) -> f₂(y, x)

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