Problem
Let $ R \subseteq A \times B $ and $ S,T \subseteq B \times C $. Proof following for combined binary relation or show that statement is false
$$ (R \circ S) \cap (R \circ T) \subseteq R \circ (S \cap T) $$
Attempt to proof
Let $ \langle x,y \rangle \in (R \circ S) \cap (R \circ T) $
$$ \iff (\langle x, y \rangle \in R \circ S) \wedge (\langle x ,y \rangle \in R \circ T) $$
$$ \iff (\exists w: \langle w,y \rangle \in S) \wedge (\exists z : \langle z, y \rangle \in T) $$
$$ \iff (\langle x , w \rangle \in R) \wedge (\langle x , z \rangle \in R) $$ $$ \iff \langle x , y \rangle \in R \circ (S \cap T) $$
Is my proof correct? I'm not quite sure if this is right.
EDIT:
Incase using $\langle x,y \rangle$ is not common notation. This means $ \langle x ,y \rangle \iff xRy $, relation $R$ maps $x$ to $y$.
Attempt to prove by counterexample
Let $ \langle x,y \rangle \in (R \circ S) \cap (R \circ T) $
$$ \iff (\langle x, y \rangle \in R \circ S) \wedge (\langle x ,y \rangle \in R \circ T) $$
$$ \iff (\exists w: \langle w,y \rangle \in S) \wedge (\exists z : \langle z, y \rangle \in T) $$
$$ \iff \{ \langle x, w \rangle, \langle x, z \rangle \} \in R $$
$$ \iff \{ \langle w, y \rangle \} \cap \{ \langle z , y \rangle \} = S \cap T = \emptyset $$
$$ \iff \langle x , y \rangle \not\in R \circ (S \cap T) $$
$$ \iff (R \circ S) \cap (R \circ T) \not\subseteq R \circ (S \cap T) $$
Is my proof by counterexample valid?