1. For Wed, read Chapter 3, section 3. Nongraded Homework: Exercises the end of the section. Even better, do Power of Logic, 7.3, A and B. Graded homework #3 is due at the beginning of class on Friday.

2. The logical connectives are truth-functions: for each possible combination of truth-values, a determinate output is specified. Represented graphically for the ampersand: p q p & q T T T T    T     Read the lines horizontally, left to right; line 1 says, “When both conjuncts are true, the entire ampersand-statement is true”.

3. In summary, an ampersand-statement is true when both sides are true; false otherwise. A vee-statement is false when both sides are false; it’s true otherwise. A tilde “reverses” the truth-value of the entire statement to which it applies. A double-arrow statement is true if and only if the truth-values of its two components match.

4. An arrow statement is false when the left side (the antecedent) is true and the right side (the consequent) is false; it’s true otherwise. From this knowledge, we can calculate the truth- value of any wff so long as we know the truth- value of its component statement letters and we understand the syntax (that is, the way in which the components are grouped together). Example: (A v ~ B)  C Say that A and B are false, and C is true.

5. Order of work: 1. Put truth-values under all capital letters 2. Apply tildes that are to the immediate left of capital letters 3. Apply tildes that are to the immediate left of tildes that are to the immediate left of capital letters (repeat as necessary) 4. Calculate truth-values for binary connectives inside the smallest sets of parentheses (those that contain no other parentheses) 5. Calculate truth-values for tildes that have only the smallest sets of parentheses in their scope 6. Repeat steps 4 and 5 for larger and larger parenthetical groupings until only the main connective has no truth-value beneath it; calculate its truth-value and you’re done—that’s the truth- value of the entire compound statement.

6. ~ [ ~ ( ~ A → B) ↔ (A & ~ D)] Let’s say that A and B are false and D is true.

7. Using truth-tables to classify formulae: First, construct a truth-table: 1. determine the number of rows needed; this is 2 n, where n = the number of distinct sentence letters in the formula; 2. write the formula horizontally across the top of the table

8. 3. To the left of the formula, list (horizontally, left to right) the distinct statement letters in the order in which they appear in the formula; put alternating Ts and  s beneath the sentence letter to the far right (closest to the premises); moving to the left, continue to fill in Ts and  s, doubling the size of the groups of Ts and  s for each new letter you come to. The last letter to the left should have 2 n /2 Ts followed by 2 n /2  s beneath it.

9. Calculate truth-values of the entire formula on each row of the truth-table. Tautology: True on every row of its truth-table (Ts all the way down underneath the main connective) Contradiction: False on every row of its truth-table (  s all the way down underneath the main connective) Contingent Statement: True on at least one row and false on at least one row of its truth-table (at least one T and at least one  underneath the main connective).

10. A B (A ↔ B) → ( ~ A & ~ B) T T T     T   T   T  T  T T    T T T T T * Answer: Contingent

11. A B A → [~ (A & B) → ~ B] T T T   T T  T  T T  T T  T T   T * Answer: Tautology

12. F G ~ (F → G) & G T T  T  T    T  T     * Answer: Contradiction

13. Using truth-tables to test for logical equivalence: Make a truth-table that includes both formulae. If their truth-values match on each row, they are logically equivalent; if there is even one line where the truth-values of the two statements (as a whole) don’t match, the statements are not logically equivalent.

14. D G ~ D v ~ G ~ (D & G) T T     T T   T T T   T T T  T    T T T T  * * * * On each line, the value of the m.o.’s match each other. So, these two statements are logically equivalent.

15. Exercises on p. 57-58