Discrete Mathematics with Graph Theory (3rd Edition) 36

# Discrete Mathematics with Graph Theory (3rd Edition) 36 -...

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34 Solutions to Exercises (b) First, we prove A U (B n G) ~ (A U B) n (A U G). So let x E Au (B nO). Then x E A or x E B n G. If x E A, then x E Au B and x E AUG, so x E (AU B) n (AUG). If x E Bn G, then x E B and x E G so x E AUB and x E AU G; that is, x E (A U B) n (A U G). In either case, x E (A U B) n (A U G) giving the desired inclusion. Second, we prove (A U B) n (A U G) ~ Au (B n G). So let x E (AUB) n (AUG). Thus, x E AUB and x E AuG. If x E A, then x E Au (BnG). If x ¢:. A, then we must have x E B and x E G; that is, x E B n G, so x E Au (B n G). In either case, x E A U (B n G) giving the desired inclusion and equality. 21. We use the fact that (XC)C = X for any set X. Let X = Nand Y = BC. Then A = XC and B = y c , so (AnB)C = [XC nycr = [(XUy)C]C (by the first law of De Morgan) = X U Y = A C U BC, as required. 22. [BB] Using the fact that X, Y = X nYc, we have (A, B) , G = (A nBC) n G C = An (B C n G C ) = An (B U G)C = A , (B U G). 23. We use the laws of De Morgan and the facts that
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