Set Difference with Intersection is Difference

Theorem

The set difference with the intersection is just the set difference.

Let $S, T$ be sets.

Then:

$S \setminus \paren {S \cap T} = S \setminus T$

Proof

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

\(=\)

\(\ds \paren {S \setminus S} \cup \paren {S \setminus T}\)

De Morgan's Laws: Difference with Intersection

\(\ds \)

\(=\)

\(\ds \O \cup \paren {S \setminus T}\)

Set Difference with Self is Empty Set

\(\ds \)

\(=\)

\(\ds S \setminus T\)

Union with Empty Set

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.1$: Theorem $1.6$
• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.6$. Difference and complement: Example $20$
• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.3$