Originally Posted by

**Bashyboy** I am currently working on proving these statements $\displaystyle \overline{A \cap B \cap C}$ and $\displaystyle \overline{A}\cup \overline{B} \cup \overline{C}$, by means of subsets (I've heard that is is also referred to as, "proof by double inclusion."). I am having trouble with these sort of problems, so I'd certainly appreciate the help.

Here is my work:

Suppose that $\displaystyle x \in \overline{A \cap B \cap C}$; for $\displaystyle x$ to be in the set, $\displaystyle x$ can't be in $\displaystyle A \cap B \cap C$. This means that $\displaystyle x \notin A \wedge x \notin B \wedge x \notin C$; however, by this fact, $\displaystyle x$ can be in $\displaystyle \overline{A}$, $\displaystyle \overline{B}$, or $\displaystyle \overline{C}$. Thus, when $\displaystyle x \in $$\displaystyle \overline{A \cap B \cap C}$, $\displaystyle x$ is also in $\displaystyle \overline{A}\cup \overline{B} \cup \overline{C}$, meaning that $\displaystyle \overline{A \cap B \cap C} \subseteq \overline{A}\cup \overline{B} \cup \overline{C}$

I have not commenced the second part, yet, because I feel as though there is something wrong with this, and I feel like I might have left out steps and explanations.