Equal Set Differences iff Equal Intersections/Proof 2

Proof
From Set Difference and Intersection form Partition:
 * $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$
 * $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$

whatever $R, S, T$ might be.

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

Then:

Now, we have from Set Difference with Disjoint Set:


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

and so:
 * $\paren {R \cap S} \setminus \paren {R \setminus S} = R \cap S$

and:


 * $\paren {R \cap T} \setminus \paren {R \setminus T} = R \cap T$

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:
 * $\paren {R \setminus S} \setminus \paren {R \cap S} = R \setminus S$

and:
 * $\paren {R \setminus T} \setminus \paren {R \cap T} = R \setminus T$

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