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# coco04hws03 - = AB(CD B'C(D'E = AB(C D B'C(D E =ABC ABD...

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Solution to ECSE 2610 Homework Set 3 (10~ '2./ ~'nt> ) I. (7 pts with I pt each part) Problem 4.6: (a) F=WXYZZ'+WXYZX'+WXYZW'+WXYZY' =0+0+0+0 =0 (b) F=AB+C'DE (c) F=(MNO+Q'P'N'+Q'OMP')+(PRM+MR) =MNO+Q'P'N'+MR Problem 4.10: (a) F= ~X,y,z(0,3)=ITx,y,z(I,2,4,5,6,7) (b) F=ITA,B,C(1,2,4)=~A,B,cCO,3,5,6,7) (e) F = X'(Y+Y')(Z+Z') + YZ'(X+X') = X'YZ+X'Y'Z+X'YZ'+X'Y'Z'+XYZ'+X'YZ' = ~X,y,z(0,1,2,3,6) F =ITx,y,z(4,5,7)= (X'+Y+Z)(X'+Y+Z')(X'+Y'+Z') (t) F=A'B+B'C+A=A'BC+A'BC'+A'B'C+AB'C+ABC+ABC'+AB'C' F= ~A,B,C(1,2,3,4,5,6,7)= ITA,B,cCO) ~(8 pts with 2 pts each part) Using DeMorgan's law to find the complement of the following expressions and obtain the dual of each expression directly from its complement. a) F=X'Y + X'Y'Z F' = (X'Y)'(X'Y'Z)' = (X+Y')(X+Y+Z') FD=(X'+Y)(X'+Y'+Z) b) F= (A'+B'+CD)(B+C'+D'E') F'= (A'+B'+CD)'+(B+C'+D'E')'
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Unformatted text preview: = AB(CD)'+B'C(D'E')' = AB(C'+D')+B'C(D+E) =ABC'+ABD'+B'CD+B'CE c) F=(((A+B)'+C')'+D)' F'= ((A+B)'+C')'+D = (A+B)C+D = AC+BC+D FD= A'C'+B'C'+D' d) F=AB(C'+D')+A'B'C'+CD'(A +C') F' = (AB(C'+D'))'(A'B'C')'((CD')'+A'C) = (A'+B'+CD)(A+B+C)(C'+D+A'C) FD= (A+B+C'D')(A'+B'+C')( C+ D'+AC') 4. (6 pts with 2 pts each part) Prove the equality of the following equations and then using the principle of duality to obtain their duals. a) AB + AB'C= AB + AC PROOF: AB+AB'C = AB+ABC+AB'C = AB+AC DUAL: (A+B)(A+B'+C) = (A+B)(A+C) b) XY'+ X'Z + Y'Z=XY'+ X'Z PROOF:XY'+X'Z+Y'Z = XY'+ X'Z+XY'Z+X'Y'Z = XY'+X'Z DUAL:(X+Y')(X'+Z)(Y'+Z) = (X+Y')(X'+Z) c) (B'+C)(B'+D)=B'+CD PROOF: (B'+C)(B'+D) = B'+B'D+B'C+CD = B'+CD DUAL: B'C+B'D = B'(C+D)...
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