(I) Let p, q, and r be the statements in Java Integrated QR-Code in Java (I) Let p, q, and r be the statements

(I) Let p, q, and r be the statements use javabean qr codes printer tobuild qr codes on java Microsoft SQL Server pThe given argument is (-"s). I I I I like mathematics study pass mathematics graduate. This is the same as p q q -*r r --. (-"r) .. ((q) V s -q which is certainly not valid, as the following partial truth table shows. s --- q p F 6. q I s FF FIT r T [(-r) -. r T T q] so, with p sq q F -r, we get p r v q is logica applet QRCode lly equivalent to [-(-r) V q]. q by the chain rule. We will prove b spring framework qr bidimensional barcode y contradiction that no such conclusion is possible. Say to the contrary that there is such a conclusion e. Since C is not a tautology, some set of truth values for p and q must make C false.

But if r is true, then both the premises (-p) -+ r and r V q are true regardless of the values of p and q. This contradicts C being a valid conclusion for this argument. In Latin, modus ponens means "method of affirming" and modus tollens means "method of denying".

This is a reflection of the fact that modus tollens has a negative -p as its conclusion, while modus ponens affirms the truth of a statement q.. 8. (a) p A q is true precisely when p and q are both true. 10. Exercises 2.1 1 tomcat QR . (a) {-,15, 15); (c) { , -2 }.

5i and 17 -43i. 2.(a) For examp Quick Response Code for Java le, I + i, I + 2i, I + 3i, -83. (a) {1,2}, {1,2,3}, {1,2,4}, {1,2,3,4}.

Solutions to Selected Exercises Only (c) is true. The set A contains one element, {a, bl. (d) False. 5. (a) True;. 6. (a) (0}. 7. (a) True; 8. (e) False; (h) False.

Yes it is; for example, let x = (l) and A = {1, I1)).. 9. (a) ii. fa, jsp QR b, c), [a, b, d), fa, c, d), fb, c, d) 11.

(a) 4; (b) 8;. (c) There are 2 " subsets of a set of n elements. (See Exercise 16 in Section 5.1 for a proof.

) 12. (a) False; (d) True..

14. (a) True. ( spring framework Quick Response Code -a) If C e 7P(A), then by definition of power set, C is a subset of A; that is, C C A.

(.a-) If C C A, then C is a subset of A and so, again by definition of power set, C E 7P(A). Exercises 2.

2 1. (a) A = (1,2,3,4,5,6}, B = (-l,0, 1,2,3,4,5}, C = 0,2, -21. 2.

(a) SnT =. 25), SUT = 2,5, V2,25,nr, 5,4,6, -}, (25, 25), (a), (aF2,25), (6, 2), (6, 25), (, ) (, 25)).. T x (S n T) = ( j2ee Quick Response Code (4, a2), (4, 25), (25, a),. (b) Z US = Jv2- r, 52 0, 1,-1, 2, -2, ...

.}; Z n S= (2, 5,251;. ZUT = ( 3. (a) Quick Response Code for Java (1,9,0,6,71 5. (a) {c, (a,b}}; 6.

(a) Ac = (-2, 1] 7. (a) 2n-2 8. (a, b)c = (-co, a] U [b, oc), [a, b)c (b) Mn P = 0.

(e) 0 4, l -1,22, 1, 2 ...

. ; ZnT = (4,25,6}..

(-oc, a) U [b, oc), (a,. 0o)C (- o, a] and (-oc, b]C (b, oc).. 9. (a) T C CS;. 11. (a) En P# 0. 12. (a) A3 UA 13. = A3. Region 2 repres ents (A n C) \ B. Region 3 represents A n B n C; region 4 represents (A n B) \ C. Think of listing the elements of the given set.

There are n pairs of the form (1, b), n -I pairs of the form (2, b), n - 2 pairs of the form (3, b), and so on until finally we list the only pair of the form (n, b). The answer is l+2+3+ +n= Wn(n+l)..

14. (a) A C B, by Problem 7. 15. 17. (b) Yes. Gi ven A n B = A n C and A" n B = A" n C, certainly we have A n B C C and A" n B C C so, from (a), we have that B C C.

Reversing the roles of B and C in (a), we can also conclude that C C B; hence, B = C. 20. Using the fact that X \ Y X n Yc, we have.

(A \B)\ C = (A QRCode for Java n B) n cc = A n (B n Cc) = A n (B U C)" = A \ (B U C). 23. (a) (AUBUC)c=[AU(BUC)]"=ACl(BUC)"= AcU(B"nCc)=AcnBcCc.

(AnBnC), =[AU(BnC)]c=A u(BnC)c =AU(BcUCc)=AcUB"Uc..
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