I am trying to prove $(1)$ below for irreversible processes, but I have a significant doubt. I will first lay out my attempt, and then explain my issue.
Let $\mathcal{Z}$ be any irreversible process which transforms a system between two states. Then $$\Delta_{\mathcal{Z}}S>\int_{\mathcal{Z}}\frac{\delta Q}{T} \tag{1}$$
where $\Delta_{\mathcal{Z}} S$ is the change in entropy of the system over the curve $\mathcal{Z}$.
I start by noting the Clausius inequality. For any irreversible, cyclic process $\mathcal{C}$, we have $(2)$ $$\oint_{\mathcal{C}}\frac{\delta Q}{T}\lt 0 \tag{2}$$
Let $\mathcal{P}:=\mathcal{R}\cup \mathcal{I}$ be a cyclic transformation where $\mathcal{R}$ transforms the system reversibly from state $A$ to $B$, and $\mathcal{I}$ transforms the system irreversibly from $B$ to $A$. By additivity of the line integral, we have $$\oint_\mathcal{P}\frac{\delta Q}{T}=\int_{\mathcal{R}}\frac{\delta Q}{T}+\int_{\mathcal{I}}\frac{\delta Q}{T}<0\tag{3}$$
The first term is simply the change in entropy, $\displaystyle \Delta _{\mathcal{R}}S=S(B)-S(A)$. Rearranging gives us equation $(4)$ $$\displaystyle \Delta_{\mathcal{R}}S<-\int_{\mathcal{I}}\frac{\delta Q}{T}\tag{4}$$
Here is where I don't really understand what is going on. If we divide by $-1$ and flip the inequality sign, then we have proven $(1)$ (since the LHS will give $S(A)-S(B)$). However, if we try to use another property of line integrals, that "flipping the sign" changes the order of the limits, then we should get $$\Delta_{\mathcal{R}}S<\int_{-\mathcal{I}}\frac{\delta Q}{T}$$
since $\mathcal{I}$ transforms our system from $B$ to $A$, $-\mathcal{I}$ should transform it from $A$ to $B$. But the inequality sign has the wrong direction here. I suspect what is happening is that I am applying a property of line integrals which is illegal, since this property of reversing the curve is typically proven for exact differentials, which $\delta Q/T$ for the process $\mathcal{I}$ is not.
