Propositional logic and truth tables

In the first two lectures, A, B, C, and D were four possible states. We can also read each state as a complete assignment of true or false values. Let X mean “there is a dot,” and Y mean “there is a stripe.” Card A makes X true and Y false. B makes both true. C makes X false and Y true. D makes both false. These four rows exhaust every possible way two true-or-false variables can be assigned. This is our running two-variable universe. Later, whenever we use a different number of variables, we will explicitly rebuild and label the carrier. A variable such as X is a simple formula. At each complete assignment, it has a value: true or false. X is true at A and B, the rows with a dot. Y is true at B and C, the rows with a stripe. A formula can be true in some states and false in others. The set of states where it is true is its truth region. For X, that region is A and B. A formula is the expression we write. Its value is true or false at one particular row. Keep those roles distinct. Now combine two formulas. “X and Y” requires both statements to be true in the same row. Which card has both a dot and a stripe? Only B. So X and Y is true at B and false at A, C, and D. The wedge symbol means “and.” Its truth region is the intersection of the truth regions for X and for Y. “X or Y” is true when at least one of the two statements is true. A qualifies because it has a dot. B qualifies because it has both. C qualifies because it has a stripe. D is the only row where neither statement holds. So the truth region of X or Y is A, B, and C. The vee symbol means “or.” Here it is inclusive: B still counts even though both statements hold. Negation changes true to false and false to true. “Not X” is true exactly where X is false. Since X means dot, not X is true at C and D: the two rows with no dot. The symbol at the front is read “not.” It is a logical operation on a complete assignment. One important reminder: this drawing describes the truth region of not X. It does not say that an observer has obtained a certificate of no dot. Description and acquired evidence are different jobs. The connective that most often surprises people is implication: “if X, then Y.” It is false in exactly one situation: X is true but Y is false. Card A is that counterexample. It has the dot required by X, but lacks the stripe required by Y. B makes the implication true: whenever X holds there, Y holds too. C and D also make the implication true. In those rows X is false, so neither supplies the forbidden combination, X true and Y false. This does not establish Y: Y is true at C and false at D. This is a rule about truth values, not a claim that a dot causes a stripe. The implication has one forbidden pattern: true X together with false Y. A truth table is simply a complete map. Each row is a possible assignment, and each formula column records its value at that row. Before we move on, look at the implication column. Which row is false, and why? A is the only false row, because it alone has X true and Y false. The table evaluates a formula at every complete possibility. It does not select the actual card or report an observation. Some formulas have a special pattern across the whole table. “X or not X” is true in every row. A formula true at every complete assignment is called a tautology. “X and not X” is false in every row. A formula false at every complete assignment is called a contradiction. Their truth regions are the whole universe and the empty set. These are logical facts about the table, before we add any observations or premises. For a fresh check, evaluate Y and not X. Which states make this compound formula true? Use both requirements at the same valuation, and keep the outer carrier fixed. Not X is true at C and D. Y is true at B and C. To satisfy their conjunction a valuation must belong to both regions. Only C belongs to both. At C, Y is true and X is false. At D, not X is true but Y is false, so D does not satisfy the conjunction. This new calculation uses the same truth-table rules, rather than repeating the implication counterexample. Every formula gives a truth region. Next we will translate those regions into set operations and compare formulas across all valuations.