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:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(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:

ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
ACK(s(x), y) → F(x, x)
F(x, y) → ACK(x, y)
ACK(s(x), 0) → ACK(x, s(0))

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
ACK(s(x), y) → F(x, x)
F(x, y) → ACK(x, y)
ACK(s(x), 0) → ACK(x, s(0))

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

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

ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
ACK(s(x), y) → F(x, x)
ACK(s(x), 0) → ACK(x, s(0))
F(x, y) → ACK(x, y)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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