📂Set Theory

Mathematical Logic Proof of Syllogism

Law ¹

The following hypothetical proposition is called a syllogism. $$ ( p \to q ) \land ( q \to r ) \implies p \to r $$

Explanation

There is hardly anyone who does not know about the syllogism, and it is believed there’s no need for explicit explanation. Except in ancient philosophical debates, it’s rare to use the phrase ‘by syllogism.’ That’s because it is a familiar method of reasoning and a universally valid principle to all of us.

However, few people would have thought that syllogism needs to be proven or could be proven. Let’s prove the syllogism using the logical symbols used in mathematics.

Proof

Logical equivalence of the conditional: $$ p \to q \equiv \left( \lnot p \lor q \right) \qquad \cdots \star $$

Various tautologies:

• [2] Simplification law: $$ p \land q \implies p \\ p \land q \implies q $$
• [7] Association law: $$ (p \land q) \land r \iff p \land (q \land r) \\ (p \lor q) \lor r \iff p \lor (q \lor r) $$
• [8] Distribution law: $$ p \land (q \lor r) \iff (p \land q) \lor (p \land r) \\ p \lor (q \land r) \iff (p \lor q) \land (p \lor r) $$

Therefore, because of $( p \to q ) \iff (\lnot p \lor q)$ $$ \begin{align*} & ( p \to q ) \land ( q \to r ) \\ \iff & (\lnot p \lor q) \land (\lnot q \lor r) & \because \star \\ \iff & (\lnot p \land \lnot q) \lor (\lnot p \land r) \lor (q \land \lnot q) \lor (q \land r) & \because [7], [8] \\ \implies& \lnot p \lor \lnot p \lor c \lor r & \because [2] \\ \implies& \lnot p \lor r \\ \implies& p \to r & \because \star \end{align*} $$ Here, $c$ represents a contradiction.

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