If I am thinking in terms of a idealized, perfect carnot cycle I know that in sum
$$\Delta S_{\mathrm{total}} = 0.\tag{1}$$
But that does not mean that there is no entropy generated during the individual steps
$$\Delta S = \frac{Q}{T}.\tag{2}$$
It is just that in total the entropy terms 'cancel out' such that equation (1) holds.
Yet when I read about other processes sometimes they seem to imply that 'quasi-static' always means $\Delta S =0$.
For example as I was reading about the Jarzynski equality he states at the very beginning of the paper:
When the parameters are changed infinitely slowly along some path γ from an initial point A to a final point B in parameter space, then the total work W performed on the system is equal to the Helmholtz free energy difference ∆F between the initial and final configurations.
$$W = \Delta F \tag{3}$$
Now this means that we have a system which moves (slowly) from point A to point B in the parameter space. The only way I can think of to obtain eq. (3) is to use the definition of the Helmholtz free energy
$$F = E - TS\tag{4}$$
re-write it in terms of
$$\Delta F = \Delta E - T\Delta S\tag{5}$$
and iff $\Delta S =0$ then I obtain $\Delta F = \Delta E$ where additionally $\Delta E = W$ because we are in a thermally isolated system yielding the solution of eq. (3).
So my question: Does the fact that we do a quasi-static, slow evolution automatically imply that $\Delta S =0$.
Because literally in the next paragraph Jarzyski says that if the evolution is not quasi static we obtain
$$W \geq \Delta F\tag{6}$$
which (assuming my investigation was correct) would mean that here $\Delta S \geq0.$