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Already read: $\wedge,\cap$ and $\vee,\cup$ between Logic and Set Theory always interchangeable?

I am learning logic for the first time, about six months after finishing my undergraduate degree. I notice that there seem to be some similarities between set theory and logic.

For example, if $A$ is a set, $a$ is a statement form, and $\mathbf{t}$ and $\mathbf{c}$ are a tautology and contradiction respectively, letting $U$ be a universal set [yes, I know this leads to a paradox] and $\varnothing$ be the empty set, I find that $U$ and $\mathbf{t}$ have similar properties, as do $\mathbf{c}$ and $\varnothing$.

Is my hunch wrong? For example, $A \cap \varnothing = \varnothing$, and $a \wedge \mathbf{c} = \mathbf{c}$.

Clarinetist
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    It seems to me that Asaf's answer in the linked question explains almost everything you want to know (that is: your intuition is correct). Of course you should take a look at Boolean algebra's definition (http://en.wikipedia.org/wiki/Boolean_algebra_(structure)) and at Stone's Representation theorem. If there is something which is still not clear to you, then you should be more precise in your question.

    In particular, observe that, by definition, in any Boolean algebra there exist two "special" elements: $0$ and $1$ (sometimes called also "top" and "bottom").

    – aerdna91 Dec 14 '14 at 22:29
  • @aerdna91 - Thank you! – Clarinetist Dec 14 '14 at 22:32
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1 Answers1

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Yes. In fact, for any set $U$, you can consider its boolean algebra of subsets. Then, when you're speaking about $U$ specifically, it is quite reasonable (and rather traditional!) to define "proposition" to mean "subset of $U$", and then define $\bot,\top,\vee,\wedge$ to mean $\varnothing,U,\cup,\cap$ and so forth.

When we're working within $U$, $U$ is the 'universal set'.

This can even be extended to typed first-order logic; from the right point of view, you can even view some kinds of set theory as being basically the same thing as some kinds of higher order logic.