MATH 211 Exam, Spring 2008

AMERICAN UNIVERSITY OF BEIRUT Math 211- Spring 2007 V/ Drop Quiz 3, section 2, March 15, 2007 fs Names oo... ccc cece cc ce cea eeeenesuesueueeeenees 6 P AUB ID number: ....... cee Section: ....... Question 1: Express the following statement using predicates, quantifiers and logical connectives, if necessary: “The e-mail address of every user can be retrieved whenever the archive contains Lt { 1) at least one message sent by every user on the system” * all users Mt abl nee Ser ga Alm). archive coutams rmedsar de mn Sujit veer wo pent mecsede an [ve dm ( Kim) a S (ums) | => tu, Rv) Question 2: Negate the following statement so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives): ~\ (ve (SyWzP(x, yz) A 32 yP(x, y,Z)) ) f . = A oefA( By 3 Feegi3)) Vv 7 Caa tg Pe r4i)| = Axfvy dy tPeegay v VEag Pod] Riu) 1 e-m a] aolduss ol(it Can tae re bre veol Question 3: Let A, B and C be sets. Explain why (Ax B)x(Cx D) and Ax(BxC)xD are not the same, | Elements oO K(CxDB) are oredoned pairs af the fore (2,0) ea) Ferme 9 Pex (Bache are S-tupis Sh the Lormm( a 5(b,2) 18) Se ; (Ax8) x (cd) s+ Ax (x0) xD Question 4: Let A, B and C be sets. A®C=BOC => A=B? uw Show that: o . Wow if equivatenr +e Show, AER =p A@C + &8® 2) (eon fra pose aq ow B Assome Ae en \ x48 ) Vv (Axrew | xen) it means | Set us assome SEK (xB — (wien ss C8) Wwe have +o show that AOC + BOC Cace kh» x€eec) , AO bur xe \ “e xe NEG and ae Cc] so > v yu Died A@C = kee => A=B AOC = BOC [A <3 | Sub-ascumehen! ac ant Ory ac €F ge 4 =xeEc ) oc ¢ NOC x E BOT bor nee 408 < é€B Cse2: oh c mE OC x E€ROC adc q wes TR beth cotet ER So fee, xek —>~esR| = ter BEA SuUb-ass Ure phow ‘ Gwdis weEec xv ar be Wea dn R = x 4g BOC ad ror} me ER xeEc | Case 2» ER LE BOC x ee Roth cagess2e€ Pe n ¢ ay MER Be Co « Auwion', ° ue a (= “> e A CB A=