Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of Set Theory Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of set theory:$$X\setminus(A\cap B)=(X\setminus A)\cup(X\setminus B)$$
 A: We have:
$x\in X\setminus (A\cup B)\Leftrightarrow x\in X\wedge (x\not\in A\cup B)\Leftrightarrow x\in X\wedge (x\not\in A\vee x\not \in B) \Leftrightarrow (x\in X\wedge x\not\in A)\vee (x\in X\wedge x\not\in B) \Leftrightarrow x\in (X\setminus A)\cup (X\setminus B)$.
Please note the use of some laws from propositional calculus, like distributivity and De Morgan's law.
A: Sketch of the proof. We want to prove that two sets are equal. Recall the definition of equality between sets.

Definition. Let $A$ and $B$ be any sets. We say that the sets $A$ and $B$ are equal, and we write $A = B,$ if $A \subseteq B$ and $B \subseteq A.$

Hence, we must prove two inclusions, by proving that for an arbitrary object $c$
\begin{align*}
c \in X\setminus(A\cap B) \implies c \in (X\setminus A)\cup(X\setminus B) && \text{and}\\
c \in (X\setminus A)\cup(X\setminus B) \implies c \in X\setminus(A\cap B).
\end{align*}

Proof. Let $x$ be an arbitrary object. Then,
\begin{align*}
x \in X \setminus (A \cap B) & \iff x \in X \wedge x \notin A \cap B & \text{(definition of set difference)}\\
& \iff x \in X \wedge \sim (x \in A \wedge x \in B) & \text{(definition of set intersection)}\\
& \iff x \in X \wedge (x \notin A \vee x \notin B) & \text{(De Morgan’s Law)}\\
& \iff (x \in X \wedge x \notin A) \vee (x \in X \wedge x \notin B) & \text{(distributive property of $\wedge$)}\\
& \iff x \in X \setminus A \vee x \in x \in X \setminus B & \text{(definition of set difference)}\\
& \iff x \in (X \setminus A) \cup (X \setminus B) & \text{(definition of set union)}
\end{align*}
Hence, we have prove that each of the sets $X \setminus (A \cap B)$ and $(X \setminus A) \cup (X \setminus B)$ is a subset of the other. Then, by definition of equality between sets we conclude that $X \setminus (A \cap B) = (X \setminus A) \cup (X \setminus B).$ $\square$

Remark. Note that in the above proof we only deal with equivalences and not just with implications, that’s because we are relying on the definitions to justify all of our steps plus the De Morgan’s Law which is itself an equivalence.
