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Why Read This Book? - Index of

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1.1 Introduction to Logic 17<br />

example, consider that p ∧ q should certainly be viewed as having the same meaning<br />

as q ∧ p. To build a truth table would produce identical columns for p ∧ q and<br />

q ∧ p. <strong>This</strong> is the way we define logical equivalence.<br />

Definition 1.1.8 Two statements are said to be logically equivalent if they have<br />

precisely the same truth table values. If U and V are logically equivalent, we write<br />

U ⇔ V .<br />

Notice that the two statements in Exercise 1.1.6 are logically equivalent. To say<br />

“p AND either q or r” has the same meaning to us as “p and q,ORp and r.”<strong>This</strong> is<br />

a sort <strong>of</strong> distributive property, where ∧ distributes over ∨, just like multiplication<br />

distributes over addition in the real numbers.<br />

EXERCISE 1.1.9 Show that ∨ distributes over ∧ by showing that p ∨ (q ∧ r) is<br />

logically equivalent to (p ∨ q) ∧ (q ∨ r).<br />

Parts (a) and (b) <strong>of</strong> Exercise 1.1.7 show that (p ∨ q) ∨ r is logically equivalent to<br />

p ∨ (q ∨ r). We say that ∨ has the associative property, and we may allow ourselves<br />

the freedom to write p ∨ q ∨ r to mean either (p ∨ q) ∨ r or p ∨ (q ∨ r).<br />

EXERCISE 1.1.10 Does ∧ have the associative property? Verify your answer<br />

with a truth table.<br />

EXERCISE 1.1.11 Construct a statement using only p, q, ∧, ∨, and ¬ that is<br />

logically equivalent to p ˙∨ q. Demonstrate logical equivalence with a truth table.<br />

EXERCISE 1.1.12 Below are two logical equivalences called DeMorgan’s laws<br />

(a name you will want to remember). Verify these forms <strong>of</strong> DeMorgan’s laws with<br />

truth tables.<br />

(a) ¬(p ∧ q) ⇔¬p∨¬q (b) ¬(p ∨ q) ⇔¬p∧¬q With Exercises 1.1.7(a) and (c), 1.1.10, and 1.1.12(a), we can extend one form<br />

<strong>of</strong> DeMorgan’s law by using the following manipulation <strong>of</strong> logical symbols:<br />

¬(p ∧ q ∧ r) ⇔¬[(p ∧ q) ∧ r] ⇔¬(p ∧ q) ∨¬r<br />

⇔ (¬p ∨¬q) ∨¬r ⇔¬p∨¬q∨¬r (1.7)<br />

EXERCISE 1.1.13 Mimic the manipulation in (1.7) to show that ¬(p ∨ q ∨ r) is<br />

logically equivalent to ¬p ∧¬q ∧¬r.<br />

EXERCISE 1.1.14 Use DeMorgan’s laws and symbolic manipulation to show<br />

that ¬[(p ∨ q) ∧ r] is logically equivalent to (¬p ∧¬q) ∨¬r.

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