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The Timpson Lyceum · First-Order Predicate Logic (2024)

Solving Exercise Problems - Lecture 4 (Logic)

Overview

This session is a problem-solving workshop that applies the machinery built up so far in the course: the rules of derivation, the notion of a truth-functional connective, tautologies, tautological implication and equivalence, sentential interpretation, and indirect (reductio) proof. Rather than introducing new theory, it works a long string of textbook exercises and shows how to attack each type. The goal is to make the foundations automatic — recognizing when a connective is truth-functional, translating English into symbolic form, deciding whether a sentence is a tautology, telling a valid argument from a fallacy, and constructing both direct and indirect derivations. A recurring lesson is that when an argument is not valid, the right move is to produce a concrete sentential interpretation making every premise true and the conclusion false.

Key Concepts

The four rules of derivation (recap). Every derivation keeps track of which premises each line depends on.

Truth-functional connectives. A connective is truth-functional when the truth value of the compound depends only on the truth values of its components — nothing else. Test: if feeding in components of the same truth value can yield different outputs, the connective is not truth-functional. Non-truth-functional examples:

English-to-symbolic translation rules.

Tautology / contradiction.

Tautological implication vs. validity. AA tautologically implies BB iff every interpretation that makes AA true also makes BB true. An argument is valid iff its premises tautologically imply its conclusion. To show invalidity, exhibit a sentential interpretation making all premises true and the conclusion false.

Sentential interpretation. Replace atomic sentences with sentences (the same sentence at every occurrence), possibly adding substructure. Rules: the main/major connectives of the original must be preserved (you may add structure, never remove it); the number of distinct atomics may increase or decrease; the count of connective occurrences may change; but the number of (not necessarily distinct) atomic occurrences can only stay the same or grow.

Common fallacies.

Two criteria for a rule of logic. Criterion 1: the rule may yield only true conclusions from true premises (soundness). Criterion 2: the rules together must be able to derive everything that follows from the premises (completeness). "From PP derive Q∨¬QQ \vee \neg Q" satisfies Criterion 1 but violates Criterion 2; "from PP derive QQ" (P→QP \to Q as a rule) satisfies Criterion 2 but violates Criterion 1.

Worked Example

Direct derivation. Premises: P→WP \to W,   P∨C\;P \vee C,   C→¬I\;C \to \neg I,   I\;I. Prove WW.

Contrapositive of C→¬I:I→¬CFrom I and I→¬C:∴ ¬CFrom P∨C and ¬C (disjunctive syllogism):∴ PFrom P and P→W (modus ponens):∴ W\begin{aligned} &\text{Contrapositive of } C \to \neg I: && I \to \neg C \\ &\text{From } I \text{ and } I \to \neg C: && \therefore\ \neg C \\ &\text{From } P \vee C \text{ and } \neg C\ (\text{disjunctive syllogism}): && \therefore\ P \\ &\text{From } P \text{ and } P \to W\ (\text{modus ponens}): && \therefore\ W \end{aligned}

Indirect (RAA) derivation. Premises: D→WD \to W,   A∨¬W\;A \vee \neg W,   ¬(D∧A)\;\neg(D \wedge A). Prove ¬D\neg D.

Assume (RAA):DRewrite premise 3 as a disjunction:¬D∨¬AFrom D and D→W:WFrom W and A∨¬W:AFrom D and ¬D∨¬A:¬AContradiction:A∧¬ADischarge the assumption, by RAA:∴ ¬D\begin{aligned} &\text{Assume (RAA):} && D \\ &\text{Rewrite premise 3 as a disjunction:} && \neg D \vee \neg A \\ &\text{From } D \text{ and } D \to W: && W \\ &\text{From } W \text{ and } A \vee \neg W: && A \\ &\text{From } D \text{ and } \neg D \vee \neg A: && \neg A \\ &\text{Contradiction:} && A \wedge \neg A \\ &\text{Discharge the assumption, by RAA:} && \therefore\ \neg D \end{aligned}

Showing an argument is invalid. For the bad argument F→SF \to S,   S→N\;S \to N,   N\;N,   ∴F\;\therefore F, interpret FF as "2+2=52+2=5", SS as "2+2=42+2=4", NN as "1+1=21+1=2". Then NN is true, both conditionals are true (true →\to true, false →\to true), yet FF is false — so the conclusion does not follow.

Summary