Answer
We argue directly by showing that if the hypothesis is true, then so is the conclusion. An alternative approach, which we show only for these parts
Work Step by Step
(a),is to use the equivalences listed in the section and work symbolically.
a)Assume the hypothesis is true. Then p is false. Since p∨q is true, we conclude that q must be true.Here is a more“algebraic” solution:[¬p∧(p∨q)]→q≡¬[¬p∧(p∨q)]∨q≡¬¬p∨¬(p∨q)]∨q≡p∨¬(p∨q)∨q≡(p∨q)∨¬(p∨q)≡T.
The reasons for these logical equivalences are, respectively, Table7, line1; DeMorgan’slaw; double negation; commutative and associative laws; negation law.
b)We want to show that if the entire hypothesis is true, then the conclusion p→r is true. To do this, we need only show that if p is true, then r is true. Suppose p is true. Then by the first part ofthe hypothesis, we conclude that q is true. It now follows from these cond part of the hypothesis that r is true, as desired.
c)Assume the hypothesis is true. Then p is true, and since these cond part of the hypothesis is true, we conclude that q is a ls o true, as desired.
d) Assume the hypothesis is true. Since the first part of the hypothesis is true, we know that either p or q is true. If p is true, then this cond part of the hypothesis tells us that r is true; similarly, if q is true, then the third part of the hypothesis tells us that r is true. Thus in either case we conclude that r is true.
