Equal Set Differences iff Equal Intersections/Proof 2

Theorem

$R \setminus S = R \setminus T \iff R \cap S = R \cap T$

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:

\(\ds \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \setminus S}\)

\(=\)

\(\ds \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \setminus T}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {R \cap S} \setminus \paren {R \setminus S}\)

\(=\)

\(\ds \paren {R \cap T} \setminus \paren {R \setminus T}\)

Set Difference with Union is Set Difference

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

\(\ds \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \cap S}\)

\(=\)

\(\ds \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \cap T}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {R \setminus S} \setminus \paren {R \cap S}\)

\(=\)

\(\ds \paren {R \setminus T} \setminus \paren {R \cap T}\)

Set Difference with Union is Set Difference

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:

$R \setminus S = R \setminus T$

$\blacksquare$