If $T ◦ R$ and $T ◦ S$ are disjoint, then so are $R$ and $S$.

Question

Suppose $R$ and $S$ are relations from $A$ to $B$ and $T$ is a relation from $B$ to $C$. If $T ◦ R$ and $T ◦ S$ are disjoint, then so are $R$ and $S$.

I am having trouble proving it (and so far, I think the above statement is true). My proof thus far is:

1. $T ◦ R \cap T ◦ S = \emptyset $
2. Then $\forall (a,c)\in A\times C ((a,c) \in T ◦ R \rightarrow (a,c) \notin T ◦ S )$
3. Suppose for contradiction $R \cap S \neq \emptyset$
4. Then $\exists(a,b) \in A \times B ((a,b) \in R \wedge (a,b) \in S) $

I'm not sure how to proceed with the proof by contradiction (not knowing whether the above statement is in fact true) since I need a $(b,c) \in T$ to use $2.$

Answer 1

It is not clear from your question if the disjointness condition for $T \circ R$ and $T \circ S$ is for any $T$ or a specific $T$. For the latter case, the statement is not true. For example, take $A=B=C=\{1,2,3\}$ and let \begin{align*} R&:=\{(1,1), (2,2)\}\\ S&:=\{(2,2), (3,3)\}\\ T&:=\{(1,1), (3,3)\} \end{align*} Then, \begin{align*} T\circ R&=\{(1,1)\}\\ T\circ S&=\{(3,3)\}\\ \end{align*} So, $T \circ R$ and $T \circ S$ are disjoint, whereas $R \cap S \neq \emptyset$.