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 $1.2.3$
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