Exercise 183Logic, sets and proofs › Sets and mapsL1Added 31 Oct 2026

De Morgan's law for sets

#sets#proofs

Statement

Let $A$ and $B$ be subsets of a set $E$ . Prove that
$E\setminus(A\cup B)=(E\setminus A)\cap(E\setminus B).$

Hint

Take an element $x$ of $E$ and translate " $x\notin A\cup B$ " using logic.

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Solution

Let $x\in E$ . Then
$x\in E\setminus(A\cup B)\iff x\notin A\cup B\iff \text{not}\,(x\in A\text{ or }x\in B)\iff x\notin A\text{ and }x\notin B.$

The last statement means $x\in E\setminus A$ and $x\in E\setminus B$ , that is $x\in(E\setminus A)\cap(E\setminus B)$ . The two sets have the same elements, so they are equal.

De Morgan's law for sets

Statement

Solution

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