Is an infinitesimal quasi-static process always reversible? I am reading Callen's Thermodynamics and introduction to Thermostatics (second edition).
In section $2.1$, Callen states that for a quasi-static process, we have 
$$\delta Q = T d S \tag{1} $$
where a quasi-static process is defined as an (ideal) succession of equilibrium states.
Now, I know (see for example Fermi, Thermodynamics) that only for a reversible process between $A$ and $B$
$$\Delta S_{A\rightarrow B} = \int_A^B \frac{\delta Q}{T}  \tag{2}$$
Since $(1)$ is basically an infinitesimal version of $(2)$, this would seem to imply that $\delta Q = T dS$ is only true for a reversible process.
Now, I know for a fact that Callen is not using the terms "quasi-static" and "reversible" interchangeably (see chapter $4$ in his book). Therefore, the only plausible conclusion I can think about to avoid a contradiction is that an infinitesimal, quasi-static process must also be reversible. Note that I am aware that in general quasi-static $\nRightarrow$ reversible (whereas reversible $\Rightarrow$ quasi-static), but I am asking about the particular case of an infinitesimal process.
Another possibility is that $(2)$ does not imply that the infinitesimal version $\delta Q= T dS$ is only valid for a reversible process, but I don't see how this could be.
Another way to rephrase my question (though slightly different) is: what is the condition to write $\delta Q = T dS$? That the process is reversible or that it is quasi-static? Or are these two conditions equivalent for an infinitesimal process?
I am aware that this could seem like an artificial problem, but I am trying to re-learn thermodynamics following an axiomatic approach, so it is very important for me to get all these conceptual issues right.
 A: The condition to write $\delta Q = T dS$ is that the process be reversible. In general, we have
$$\delta Q \leq T dS.$$
A quasi-static process is not generally reversible. Taking infinitesimals makes no difference. Indeed, if you thought that every infinitesimal quasi-static process were reversible, you would have to conclude that all quasi-static processes were reversible, since by definition every quasi-static process is composed of a series of infinitesimal steps.
For a concrete example, consider two gases separated by an insulating piston with friction. One can consider a quasi-static process where the piston slowly moves to the right. Even an infinitesimal motion is not reversible, because heat is produced by friction. The equality $\delta Q = T dS$ is violated, as $dS >0$, while $\delta Q = 0$.
I agree that it's frustrating that sources disagree on this. Part of the issue is that whenever you're doing something concrete, it's easy to see what to do (i.e. you certainly wouldn't forget the friction in the above example). That leads to a degree of sloppiness when formulating general principles, and inconsistency between and within sources. My impression from answering thermodynamics questions on this site is that Callan is sloppier than most textbooks.
A: I think that for many authors (especially the old one) quasi-static <=> reversible. Fermi that you cited says:

A transformation is said to be reversible when the successive states
  of the transformation differ by infinitesimals from equilibrium
  states. A reversible transformation can therefore connect only those
  initial and final states which are states of equilibrium. A reversible
  transformation can be realized in practice by changing the external
  conditions so slowly that the system has time to adjust itself
  gradually to the altered conditions.

For him the concepts are equivalent. If you think about it the only difference is when friction or similar processes are involved. The equation should be valid for a quasi-static transformation.
