Solve this :  Prove using properties of sets and their complements .

( 1 )   A B A B ' = A ( 2 )   A - A B = A - B ( 3 )   A B - C = A - C B - C ( 4 )   A - B C = A - B A - C ( 5 )   A B - C = A B - A C .


 
 

Dear student
(3) AB-C=ABC'=(AC')(BC')=(A-C)(B-C) (4) A-BC=A-BA-CLHS=A-BC=ABC'     X-Y=XY'=A(B'C')     [(BC)'=B'C']=(AB')(AC')=(A-B)(A-C)=RHS(5)  A(B-C)=(AB)-ACLet x be any arbitrary element of A(B-C).ThenxA(B-C)xA and xB-CxA and xB and xCxA and xB and xA and xCx(AB) and xACx(AB)-ACSo, A(B-C)(AB)-ACSimilarly, (AB)-ACA(B-C)Hence,A(B-C)=(AB)-ACSimilarly try remaining parts.
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