I'm looking at the following expression:
$$ (\forall a \in A, c \in C: (\exists b \in B: (a, b) \in R \,\land\,(b,c) \in S_1 \Leftrightarrow \exists b \in B: (a, b) \in R \,\land\, (b, c) \in S_2)) \implies S_1 = S_2 $$
where $A, B, C$ are sets ($R \subseteq A \times B, S_1, S_2 \subseteq B \times C$). My gut tells me this expression simplifies to a really simple statement about $R$, but I can't seem how to prove this.
This seems to tell that $$R\circ S_1 = R\circ S_2 \;\Longrightarrow\; S_1=S_2.$$ Let $$\pi_1(R)=\{a \in A : \exists b \in B \;((a,b) \in R)\},$$ $$\pi_2(R)=\{b \in B : \exists a \in A \;((a,b) \in R)\},$$ and similarly define $\pi_i(S_j)$ ($i,j = 1,2$).
The above implication holds if $\pi_2(R)\supseteq \pi_1(S_1)\cup\pi_1(S_2)$; in particular, if $\pi_2(R)=B.$
Edit: What could be something your gut is after, is that if $R$, $S_1$ and $S_2$ are graphs of functions, then $\pi_1(S_i)=B$, whence $\pi_2(R)=B$, that is, the function that has $R$ as its graph is surjective.