# De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection

## Theorem

Let $S, T_1, T_2$ be sets.

Then:

$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$

where:

$T_1 \cap T_2$ denotes set intersection
$T_1 \cup T_2$ denotes set union. ### Corollary

Suppose that $T_1 \subseteq S$.

Then:

$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)$

## Proof

 $\displaystyle$  $\displaystyle x \in S \setminus \paren {T_1 \cap T_2}$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in S} \land \paren {x \notin \paren {T_1 \cap T_2} }$ Definition of Set Difference $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in S} \land \paren {\neg \paren {x \in T_1 \land x \in T_2} }$ Definition of Set Intersection $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in S} \land \paren {x \notin T_1 \lor x \notin T_2}$ De Morgan's Laws: Disjunction of Negations $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in S \land x \notin T_1}) \lor \paren {x \in S \land x \notin T_2}$ Rule of Distribution $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$ Definition of Set Union and Definition of Set Difference

By definition of set equality:

$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$

$\blacksquare$

## Source of Name

This entry was named for Augustus De Morgan.