Equal Set Differences iff Equal Intersections

Theorem

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

Proof 1

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle R \setminus S\)	\(=\)	\(\displaystyle R \setminus T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left\{ {x \in R: x \notin S}\right\}\)	\(=\)	\(\displaystyle \left\{ {x \in R: x \notin T}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Set Difference
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \forall x \in R: \quad x \notin S\)	\(\iff\)	\(\displaystyle x \notin T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \forall x \in R: \quad x \in S\)	\(\iff\)	\(\displaystyle x \in T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left \{ {\left({x \in R}\right) \land \left({x \in S}\right)}\right\}\)	\(=\)	\(\displaystyle \left \{ {\left({x \in R}\right) \land \left({x \in T}\right)}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle R \cap S\)	\(=\)	\(\displaystyle R \cap T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Set Intersection

$\blacksquare$

Proof 2

From Set Difference and Intersection form Partition that:

• $\left({R \setminus S}\right) \cup \left({R \cap S}\right) = R = \left({R \setminus T}\right) \cup \left({R \cap T}\right)$
• $\left({R \cap S}\right) \cap \left({R \setminus S}\right) = \varnothing = \left({R \cap T}\right) \cap \left({R \setminus T}\right)$

whatever $R, S, T$ might be.

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\left({R \setminus S}\right) \cup \left({R \cap S}\right)}\right) \setminus \left({R \setminus S}\right)\)	\(=\)	\(\displaystyle \left({\left({R \setminus T}\right) \cup \left({R \cap T}\right)}\right) \setminus \left({R \setminus T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({R \cap S}\right) \setminus \left({R \setminus S}\right)\)	\(=\)	\(\displaystyle \left({R \cap T}\right) \setminus \left({R \setminus T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Difference with Union is Set Difference

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 $R \cap S = R \cap T$ :

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\left({R \setminus S}\right) \cup \left({R \cap S}\right)}\right) \setminus \left({R \cap S}\right)\)	\(=\)	\(\displaystyle \left({\left({R \setminus T}\right) \cup \left({R \cap T}\right)}\right) \setminus \left({R \cap T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({R \setminus S}\right) \setminus \left({R \cap S}\right)\)	\(=\)	\(\displaystyle \left({R \setminus T}\right) \setminus \left({R \cap T}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Difference with Union is Set Difference

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:

$R \setminus S = R \setminus T$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 8 \ \text{(c)}$