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How would you convert set logic to propositional logic? In particular, I'm not sure how to handle converting $\subseteq$

For example: $$A-(\bar{B} \cup \bar{C}) \subseteq B \cap C$$

My attempt at converting to propositional logic: $$ x \in A \land (x \in B \lor x \in C) \implies x \in B \land x \in B$$

Is this the right way?

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Yes, you're correct that $\subseteq$ corresponds to the conditional connective $\rightarrow$.

But you've made an error: The antecedent in your translation corresponds to $A - (B \cup C)$, whereas we need to translate $A - (\overline B \cup \overline C)$. So we need DeMorgan's: In particular, note how the side to the left of the conditional is translated:

$$A-(\overline{B} \cup \overline{C}) \subseteq B \cap C$$

$$x\in A \land \lnot [x \in (\overline B \cup \overline C)] \rightarrow x \in B \land x \in C$$ $$\iff A \land \lnot[\lnot (x \in B\cap C)] \rightarrow x \in B \land x \in C\tag{DeMorgan's}$$ $$\iff x \in A \land (x\in B \land x \in C) \rightarrow x \in B \land x\in C$$

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  • $\begingroup$ Truly in your element! +1 :-) $\endgroup$ – Amzoti Nov 18 '13 at 0:08
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You're close, but it's not quite correct. It looks like you forgot to switch the $\cup$ to a $\cap$ (or perhaps the $\vee$ to a $\wedge$) when you applied de Morgan's law. It should be $x \in B \wedge x \in C$, not $x \in B \vee x \in C$. In general, the fast-and-loose rule is to convert $A - B$ to $A \wedge \neg B$, $A \cup B$ to $A \vee B$, $A \cap B$ to $A \wedge B$, and $\overline{A}$ to $\neg A$, using de Morgan's laws as you go. Try drawing a Venn diagram too; in this case it would be very helpful.

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