snhu MODULE TWO PROBLEM SET This document is proprietary to Southern New Hampshire University. It and the problems within may not be posted on any non-SNHU website. Zainab Lowe 1
snhu Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit. PROBLEM 1 Part 1. Indicate whether the argument is valid or invalid. For valid arguments, prove that the argument is valid using a truth table. For invalid arguments, give truth values for the variables showing that the argument is not valid. (1) (p/q) > r : (p V q) >r The argument is invalid. To show this, let's consider a specific truth value assignment for the variables p, q, and r that makes the premise true and the conclusion false: Let p = True, q = False, and r = False. In the premise (p Aq) > r, since q = False, p / q is False. In a material implication, if the antecedent (the part before the arrow) is False, the entire implication is True regardless of the truth value of the consequent (the part after the arrow). So, the premise is True. In the conclusion (p V q) > r, since p = True, p V q is True. Since r = False, the implication True -> False is False. So, the conclusion is False. Thus, there is at least one case where the premise is true and the con- clusion is false, which means the argument is invalid. Part 2. Converse and inverse errors are typical forms of invalid argu- ments. Prove that each argument is invalid by giving truth values for the variables showing that the argument is invalid. You may find it eas- ier to find the truth values by constructing a truth table. (a) Converse error p -> q q .. p
snhu To prove invalidity, let's assign truth values: Let p = False, q = True. · In p -> q, if p is False and q is True, the implication is True (since an implication is only false if the first part is true and the second part is false). · q is True as per our assignment. . The conclusion p is supposed to be True according to the argument, but in our assignment, p is False. Thus, the argument is invalid because it's possible for the premises to be true while the conclusion is false. (b) Inverse error p -> q To prove invalidity, let's assign truth values: Let p = False, q = True. . In p -> q, with p False and q True, the implication is True. · p is True, as p is False. . The conclusion -q is supposed to be True according to the argument, but in our assignment, q is True, so -q is False. Thus, this argument is also invalid because the premises can be true while the conclusion is false. Part