📂Set Theory

Mathematical Logical Proof of Syllogism

Law¹

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

Explanation

There is no one who does not know the syllogism, and it hardly seems necessary to explain it. Unless one is engaged in an ancient philosophical debate, it is uncommon to explicitly say ‘by syllogism’. That is because it is such a familiar mode of reasoning and a universally valid principle to us.

However, few people would have thought that the syllogism is something that can be proven and that should be proven. Let us 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] Associative law: $$ (p \land q) \land r \iff p \land (q \land r) \\ (p \lor q) \lor r \iff p \lor (q \lor r) $$
• [8] Distributive law: $$ p \land (q \lor r) \iff (p \land q) \lor (p \land r) \\ p \lor (q \land r) \iff (p \lor q) \land (p \lor r) $$

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

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