From the definition of entropy change,
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{int.rev}$$
From the closed system entropy balance, we have
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{b}+\sigma $$ where $\sigma$ is the entropy produced within the system, vanishing to zero in the absence of irreversibilities. I don't quite understand how the entropy change between two states is the same for all processes. Is it the case that the entropy transfer in the case of internal irreversibilities present is lower than that of the entropy transfer in an internally reversible process between these same two states, and the entropy production makes up for this difference?