Let $R$ be a relation from $A$ to $B$, and $S$ be a relation from $B$ to $C$, and $T$ be a relation from $C$ to $D$.
I want to prove that $T\circ (S\circ R)=(T\circ S)\circ R$.
This is how I proved it:
Let $(a,d)\in T\circ (S\circ R)$, so then, following the definitions, we get:
$$(a,d)\in T\circ (S\circ R)$$
$$\Leftrightarrow \exists c\in C\space s.t. [(a,c)\in S\circ R \wedge (c,d)\in T]$$
$$\Leftrightarrow \exists c\in C,\space\exists b\in B\space s.t. [(a,b)\in R\wedge (b,c)\in S \wedge (c,d)\in T]$$
$$\Leftrightarrow \exists b\in B\space s.t. [(a,b)\in R\wedge (b,d)\in T\circ S]$$
$$\Leftrightarrow (a,d)\in (T\circ S)\circ R$$
I skipped a few subtle stages that relies on the associativity that comes with logical connective AND ($\wedge$), but that's not the issue here.
There's probably an easy answer I'm overlooking, but this still bothers me:
The problem is that my proof starts with the assumption that $T\circ (S\circ R)$ is not empty.
How can I justify the equality $T\circ (S\circ R)=(T\circ S)\circ R$ for when $T\circ (S\circ R)=\phi$?
If $T\circ (S\circ R)=\phi$, that doesn't mean $T=\phi$ nor $(S\circ R)=\phi$...
To prove that two sets $A$ and $B$ are equal it is enough to show that $x\in A$ implies $x\in B$ and conversely that $x\in B$ implies $x\in A$ (as you did). If $A=\emptyset$ then $x\in B$ implies that $x\in A$ wich cannot be true. So it is okay to conclude that $B=\emptyset$.