Are zero heat exchange and zero work done synonymous to reversibility? Instead of using the definition involving entropy or quasi-static processes, if I am simply able to prove that both work done (on the gas or by the gas, doesn't really matter) and heat exchange through a process are zero, is it sufficient to prove that the process must have been reversible? If so, does it mean that $every$ process with zero heat exchange and work is a reversible one? 
 A: Consider the first law
$$ \Delta U = W + Q. $$
It is simply a restatement of conservation of energy. Thus if you force $W=Q=0$, then your system can have no change in energy, whatsoever. Nor can it make any energy exchange with the environment. 
Take the thermodynamic identity,
$$\mathrm{d}U = T \, \mathrm{d}S - P \, \mathrm{d}V + \mu \, \mathrm{d}N.$$
If we conclude that $\mathrm{d}U = 0$, then we can only make comments about the relationships here. In a case where $\mathrm{d}N = 0$, we have
$$ T \, \mathrm{d}S = P \, \mathrm{d}V, $$
so the process is only reversible if there is no change in volume. Any change in volume of the system would constitute work, so we would see that the entropy could never increase if we fix $W_{\rm net} = 0$. So you would say the process is reversible in that case.
A: We need only one counterexample to show that the hypothesis is wrong. Here is a counterexample. An insulated vessel is divided into two chambers by a partition. One chamber contains a gas, the other is evacuated. A split occurs in the partition, so that the gas expands into the chamber that used to be evacuated. This is an irreversible change (called 'Joule expansion'). Yet no work is done on or by the system, and no heat flows in or out.  
