# Increase in entropy derivation

so i'm following a derivation of the statement in the title and it goes the following way. Consider we have a system and we look at two processes between two states $$(1,2)$$ and $$(2,1)$$ where path $$(1,2)$$ is reversible and $$(2,1)$$ is irreversible. So we have: $$\oint dS=\int_{1}^{2} dS + \int_{2}^{1} dS \le 0$$ Since $$(1,2)$$ is a reversible process we can write in the following way: $$\int_{2}^{1}dS +(S_2-S_1) \le 0 \\ \Rightarrow \int_{2}^{1}dS \le (S_1-S_2)$$ $$\Rightarrow S_{irr} + \int_{2}^{1}dS = (S_1-S_2)$$ Where $$S_{irr}$$ is the bonus increase in entropy due to the irreversability of the process. So now if we say that the system is isolated $$(\delta Q=0)$$: $$\int_{2}^{1}dS=\int_{2}^{1}\frac{\delta Q}{T}=0$$ $$\Rightarrow S_{irr} = (S_1-S_2)$$ My question here is why when we assume that the system is isolated don't we say that the other integral is also going to result in zero $$(S_2-S_1=0)$$?

Most of your analysis seems incorrect to me. For the combination of reversible and irreversible cycle you describe, $$\Delta S_{cycle}=\Delta S_{1,2}+\Delta S_{2,1}=0$$That is, for a system which undergoes a cycle, the initial and final states of the overall cycle are identical, and the entropy change is zero. If the irreversible part of the cycle is adiabatic or isolated, $$\Delta S_{2,1}=\sigma>0$$where $$\sigma$$ is the entropy generation during the irreversible part of the path. So we have $$\Delta S_{1,2}+\sigma=0$$or $$\Delta S_{1,2}=\int_1^2{\frac{dQ}{T}}=-\sigma<0$$This tells us that the reversible part of the path cannot be adiabatic or isolated, and some heat transfer must occur over this part of the path. There is no possible cyclic path comprised of an irreversible- and a reversible adiabatic segments.