Equal Set Differences iff Equal Intersections/Proof 2

Proof
From Set Difference and Intersection form Partition:
 * $\left({R \setminus S}\right) \cup \left({R \cap S}\right) = R = \left({R \setminus T}\right) \cup \left({R \cap T}\right)$
 * $\left({R \cap S}\right) \cap \left({R \setminus S}\right) = \varnothing = \left({R \cap T}\right) \cap \left({R \setminus T}\right)$

whatever $R, S, T$ might be.

Let $R \setminus S = R \setminus T$.

Then:

Now, we have from Set Difference with Disjoint Set:


 * $S \cap T = \varnothing \iff S \setminus T = S$

and so:
 * $\left({R \cap S}\right) \setminus \left({R \setminus S}\right) = R \cap S$

and:


 * $\left({R \cap T}\right) \setminus \left({R \setminus T}\right) = R \cap T$

So:
 * $R \cap S = R \cap T$

We can use exactly the same reasoning if we assume $R \cap S = R \cap T$ :

and then because of Set Difference with Disjoint Set as above:
 * $\left({R \setminus S}\right) \setminus \left({R \cap S}\right) = R \setminus S$

and:
 * $\left({R \setminus T}\right) \setminus \left({R \cap T}\right) = R \setminus T$

So:
 * $R \setminus S = R \setminus T$