Prove that $( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing$. I am having a bit of a hard time proving the following statement:

Show that
  $( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing$.

I sort of understand composition of functions, inverse functions, and set theory individually, but when put together I seem to get confused.
It would be wonderful if anyone could be of help, as I'm not even sure how to begin.
Thank you!
EDIT: See attempt at the question in the answers

$( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing \iff ( R \circ S ) \cap T \ne \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$ 
Let R, S, T be relations on the same set  A


*

*$( R \circ S ) \cap T \ne \varnothing \to (\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$
Assuming $( R \circ S ) \cap T \ne \varnothing$,
Then, $\exists x, y \in A$ such that $(x,y) \in R \circ S$ and $(x,y) \in T$
$(x,y) \in R \circ S \implies \exists z$ such that $(x,z) \in R$
and $(z,y) \in S$
$(x,z) \in R \implies (z,x) \in \mathrm{R}^{-1}$
From $(z,x) \in \mathrm{R}^{-1}$ and $(x,y) \in T$, we can say that $(z,y) \in \mathrm{R}^{-1} \circ T$
Now, from $(z,y) \in \mathrm{R}^{-1} \circ T$ and $(z,y) \in S$, we can conclude that $(z,y) \in (\mathrm{R}^{-1} \circ T) \cap S$. 


*$( R \circ S ) \cap T \ne \varnothing \leftarrow (\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$
Assuming $(\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$,
Then, $\exists x, y \in A$ such that $(x,y) \in \mathrm{R}^{-1} \circ T$ and $(x,y) \in S$
$(x,y) \in \mathrm{R}^{-1} \circ T \implies \exists z$ such that $(x,z) \in \mathrm{R}^{-1}$ and $(z,y) \in T$
$(x,z) \in \mathrm{R}^{-1} \implies (z,x) \in R$
From $(z,x) \in R$ and $(x,y) \in S$, we can say that $(z,y) \in R \circ S$
Now, from $(z,y) \in R \circ S$ and $(z,y) \in T$, we can conclude that $(z,y) \in ( R \circ S ) \cap T$.
