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.

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.