Analog between Logic and Set Theory

Theorem
The concepts of set theory have directly corresponding concepts in logic:


 * {| border = "1"

! style="padding: 2px 10px" | Set Theory ! style="padding: 2px 10px" | Logic
 * align="left" style="padding: 2px 10px"| Set: $S, T$
 * align="left" style="padding: 2px 10px"| Statement: $p, q$
 * align="left" style="padding: 2px 10px"| Union: $S \cup T$
 * align="left" style="padding: 2px 10px"| Disjunction: $p \lor q$
 * align="left" style="padding: 2px 10px"| Intersection: $S \cap T$
 * align="left" style="padding: 2px 10px"| Conjunction: $p \land q$
 * align="left" style="padding: 2px 10px"| Subset: $S \subseteq T$
 * align="left" style="padding: 2px 10px"| Conditional: $p \implies q$
 * align="left" style="padding: 2px 10px"| Symmetric Difference: $S * T$
 * align="left" style="padding: 2px 10px"| Exclusive Or: $p \oplus q$
 * align="left" style="padding: 2px 10px"| Complement: $\relcomp {} S$
 * align="left" style="padding: 2px 10px"| Logical Not: $\lnot p$
 * align="left" style="padding: 2px 10px"| Set Equality: $S = T$
 * align="left" style="padding: 2px 10px"| Biconditional: $p \iff q$
 * align="left" style="padding: 2px 10px"| Venn Diagram
 * align="left" style="padding: 2px 10px"| Truth Table
 * }
 * align="left" style="padding: 2px 10px"| Logical Not: $\lnot p$
 * align="left" style="padding: 2px 10px"| Set Equality: $S = T$
 * align="left" style="padding: 2px 10px"| Biconditional: $p \iff q$
 * align="left" style="padding: 2px 10px"| Venn Diagram
 * align="left" style="padding: 2px 10px"| Truth Table
 * }
 * align="left" style="padding: 2px 10px"| Truth Table
 * }

Proof
Let $P$ and $Q$ be propositional functions.

Let $S$ and $T$ be subsets of a universe $\Bbb U$ such that:
 * $S = \set {x \in \Bbb U: \map P x}$
 * $T = \set {x \in \Bbb U: \map Q x}$

By the following definitions: