# For Wednesday, read Chapter 3, section 4. Nongraded Homework: Problems at the end of section 4, set I only; Power of Logic web tutor, 7.4, A, B, and C.

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For Wednesday, read Chapter 3, section 4. Nongraded Homework: Problems at the end of section 4, set I only; Power of Logic web tutor, 7.4, A, B, and C Graded HW#3 is due at the beginning of class on Friday.

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

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

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.

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).

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

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

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

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.

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.

Using Truth-tables to test arguments for validity. Place the entire argument on a truth-table: list the premises from left to right, separated by commas; put the conclusion on the far right after a ‘  ’  Fill out the table. If there is at least one row where all premises are true and the conclusion false, then the argument is invalid; if there is no row with all true premises and a false conclusion, the argument is valid.

Z S GZ → (S v G), Z & G  S T T T T T T T T T  T T  T T  T T T T  ***** T        T T T T  T  T  T T  T   T T T      T    * * * * * * Answer: Invalid

Why does it work? Each row on the truth-table represents a relevant possibility (an interpretation), and taken together the rows represent all of the relevant possibilities (all possible interpretations). So, if there is a row on the truth-table with all true premises and a false conclusion, then it is possible to have all true premises and a false conclusion—thus the argument is invalid; and if there is no row with all true premises and a false conclusion, it is impossible to have all true premises and a false conclusion, and by definition, the argument is valid.

P Q SP → Q, S → Q, ~ Q  ~ P & ~ S T T T T T     T T  T T    T T  T   T    T    T T   T  T T T T  T    T  T T  T T T   T T  T T      T T T T T T

Answer? Valid. There is no line showing all true premises and a false conclusion. Thus, it is impossible for the argument to have all true premises and a false conclusion. So, the reasoning can’t go wrong. Whenever the premises are all true, the conclusion is as well. Remember, when an argument is valid, there is no particular line that proves validity.

A B GA → (B & G), ~ B  A  ~ B T T T T  T T    T  T   T      T T T   T  T    T T T   T F F F Answer: Invalid, proven by line seven

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