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Unformatted text preview: PHIL V3411 / PHIL G4415 • Introduction to Symbolic Logic Lecture 4 Sentential Logic: Truthtables, Logical equivalence INTRODUCTION TO SYMBOLIC LOGIC, LECTURE 4 P. 1 A Puzzle You are on an island where there are two kinds of inhabitants: 1) Knights, who always tell the truth 2) Knaves, who always lie. You meet two of them, Alf and Beth. Alf says: At least one of us is a Knave. What are Alf and Beth? Solution Alf cannot be a Knave, since if he were, what he says would be true (which is impossible: Knaves always lie). So he must be a Knight. But if he is a Knight, then what he says is true. So the other person, Beth, must be a Knave. INTRODUCTION TO SYMBOLIC LOGIC, LECTURE 4 P. 2 Formal Explanation Alf’s statement, At least one of us is a Knave, can be expressed as the negation of a conjunction: We are not both Knights i.e. It is not the case that (Alf is a Knight and Beth is a Knight) which has the logical form ¬ ( A ∧ B ) We can construct a truthtable for this expression: A B A ∧ B ¬ ( A ∧ B ) T T T F T F F T F T F T F F F T Note: T means that Alf is a Knight F means that Alf is a Knave Inspection of the table shows that only the second row is possible (since ¬ ( A ∧ B ) corresponds to a statement of Alf’s). ) corresponds to a statement of Alf’s)....
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This note was uploaded on 10/05/2008 for the course PHIL V3411 taught by Professor Varzi during the Spring '08 term at Columbia.
 Spring '08
 Varzi

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