Termination w.r.t. Q proof of TRS/AProVErta3.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:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, 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:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
F₂(s₁(x), y) -> F₂(x, s₁(x))
F₂(x, s₁(y)) -> F₂(y, x)
ACK₂(s₁(x), y) -> F₂(x, x)
F₂(x, y) -> ACK₂(x, y)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, 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:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
F₂(s₁(x), y) -> F₂(x, s₁(x))
F₂(x, s₁(y)) -> F₂(y, x)
ACK₂(s₁(x), y) -> F₂(x, x)
F₂(x, y) -> ACK₂(x, y)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

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