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My teacher taught us mathematical logic by relating it to set theory (not entirely). He asked us to relate the propositional variable p to a set P, and similarly q to a set Q. The intuitive relations were- $$ p ∧ q\equiv P \cap Q , p ∨ q\equiv P \cup Q, p\rightarrow q\equiv \overline{P}∪Q $$

He says that we can relate the value of a propositional variable to a set like-

If p has a truth value of False then take it as an empty set and if it has a truth value of True then take it as the universal set S or U.

This literally helps in simplifying long mathematical statements easily by converting them to sets and then to converting them again to propositional variables. ( rather than writing long and hard truth tables)

Is this an easy way to learn logic or is it because it just simplifies thing fast? I want to know if this approach for simplifying things out is sound, or will i run into some roadblocks later.

Any suggestions and comments are welcome if they make this intuitive understanding more concrete.

Just a heads-up I'm just a high-school student.

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    $\begingroup$ IMO it is not "faster"... You have to rely on elementary set theory. In addition, in what sense the intuition about the empty set is easier than the intuition about False? Since our childhood we are accustomed to the difference between a false statement and a true one. $\endgroup$ – Mauro ALLEGRANZA Feb 22 at 14:18
  • $\begingroup$ @MauroALLEGRANZA, IMO and almost all my classmates it's easier to go by set theory probably because it has been taught first. But, I want to know if this approach for simplifying things out is sound, or will i run into some roadblocks later. $\endgroup$ – AnonymousDeeply Feb 22 at 14:23
  • $\begingroup$ The "common core" is called Boolean algebra $\endgroup$ – Mauro ALLEGRANZA Feb 22 at 15:05
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There is indeed a relation between logic expressions and set theory expression. For given set of basic logic sentences $p, q, \dots$ you can create space $\Omega$ of all possible possibilite situations (that is all possible combinations of logical values that your sentences may have). You can relate every logical sentence to the set of situations in which this sentence is true, that is $$ p \text{ is true in situation } x \Leftrightarrow x \in P$$ Then you have for example \begin{align} (p \land q) \text{ is true in situation } x &\Leftrightarrow (p \text{ is true in situation } x )\land (q \text{ is true in situation } x) \Leftrightarrow \\ & \Leftrightarrow (x\in P) \land (x\in Q) \Leftrightarrow \\ & \Leftrightarrow x \in (P \cap Q) \end{align} which means that the set $P \cap Q$ is the set related to the sentence $p \land q$. Analogously you can find the set operations related to other logic operations. The construction is solid, and if you already know set theory it can be useful to learning logic. For example it can visualise various logic laws (like the de Morgan's laws for example) or allow to immediately teanscribe some theorems about set on the theorems about logic sentences. For example, if you know that $$ P \cap (Q \cup R) = (P \cap Q) \cup (P \cap R)$$ then you can immediately tell that sentences $$ p \land (q \lor r) $$ and $$ (p \land q) \lor (p \land r)$$ are equivalent.

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  • $\begingroup$ that was very helpful.. we do use de Morgan's law extensively in solving questions like you have stated. Thankyou very much for clearing my doubt. $\endgroup$ – AnonymousDeeply Feb 23 at 5:26
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In some sense, the answer to your question is yes. It gives a geometric viewpoint on Boolean algebra (the algebra of propositions), known as Stone duality. A proposition is thought of as some subspace and it can be true or false depending on your position. Even if you don't learn Stone duality, this viewpoint is helpful and sound.

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  • $\begingroup$ Yes that is correct, it's the points we consider, not the proposition. $\endgroup$ – AnonymousDeeply Feb 22 at 14:37
  • $\begingroup$ @DeepJ When you write $P ⊆ Q$, you mean $\overline{P} ∪ Q$? $\endgroup$ – Idéophage Feb 22 at 14:39
  • $\begingroup$ That was an error from my side. I have got it corrected. Thks for pointing it out. $\endgroup$ – AnonymousDeeply Feb 22 at 14:41

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