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X + Y b. X + Y C. X + Y d. X + Y In Boolean algebra XY + YZ + X + Z is equal to: a. X + Z b. X + Z c. X + Z d. X + Z In Boolean algebra X+XY+YZ is equal to: a. X+YZ b. Y ZX C. Z+ XY d. X+Y+Z In Boolean algebra A C + ABC is equal to: a. AB b X C. 1 d. Z In Boolean algebra ZIzz is equal to: a. XZ b. XZ C. XZ d. XZ In Boolean algebra, the complement of (X+Y) is: a. XY b. XY C. XY d. XY In Boolean algebra F(A,B,C) = X (0, 4, 7) is same as: a. II(1,2,3,5,6) b. II(0,4,7) c. [I(0,2,4,6) d. II(1,3 Z b. Z C. X d. X In Boolean algebra XZ+Z+1 is equal to: a. 0 b. X C. 1 d. Z In Boolean algebra ZIZX is equal to: a. XZ b. XZ C. XZ d. XZ In Boolean algebra, the complement of (X+Y) is: a. XY b. XY C. XY d. XY In Boolean algebra F(A,B,C) = Y, (0,.4, 7) is same as: a. TI(1,2,3,5,6) b. II(0,4,7) c. [I(0,2,4,6) d. II(1,3,5,7) In Boolean algebra F(A,B,C) = X (1, 2, 5, 6) is same as: a. TI(1,2,5,6) b. II(0,4,6,7) c. [I(0,1,4,6) d. II(0,3,4,7) In Boolean algebra F(A,B,C) = II Simplify the following Boolean expressions by using only laws of Boolean and then construct the circuit diagrams of reduced Boolean expression (a) AB' + C'D' (b) (BC + A'D)(AB'+CD') (c) {BC + AC + AB + BCD) 3. Find the minimal form of the Boolean function of four variables represented by the Karnaugh map given by Fig. 7.41 4. Use the Karnaugh map representation to find a minimal form each of the following functions (a) f{x, y) = x'y + xy (b) f(x, y, z) = xyz + xy'z + x'yz +.x'y'z (c) f{x, (a) X + X'YZ (b) (X + KOtX' + Y) (c) (x + rxr + z) (d) XY + xz + YZ 3. Use Venn diagrams to decide which of the following are valid for all sets X, Y, and Z. (a) X(Y + Z) = XY+XZ (b) XY + X'Y + XY' = 1 (c) X(X + Y) X (d) X + XY = X (e) X' + Y' = (X + IT (f) X } X' = 1 (g) AT + X'K = X + 7 (h) X'Y' + XT + In Section 14 some of the basic identities which are valid in the algebra of sets (and in any Boolean algebra) were mentioned in connection with Venn diagrams. These laws and others which Concise text begins with overview of elementary mathematical concepts and outlines theory of Boolean algebras; defines operators for elimination, division, and expansion; covers syllogistic.reasoning, solution of Boolean equations, P. Mukhopadhyay, Shamik Ghosh, Mridul Kanti Sen. 0 Exercise 15.2.2. Consider the Boolean function f(x,y,z) given by the following table: Find the Boolean expression in CNF in x,y, z for the Boolean function /. Solution. The function / takes the value 0 for the assignments in the 4th, 7th, and 8th rows of the given table. The maxterms corresponding to these rows are (x' + y + z), (x + y + z1) and (x + y + z) respectively. Hence, the Boolean expression in CNF representing the function / is (x' (a) XY (b) XY (c) XY (d) XY 49. In Boolean algebra, if F = ( )( ) A BA C + + , then (a) F AB AC = + (b) F AB AB = + (c) F AC AB = + (d) F AA AB = + 50. The simplified form of a logic function.Y AB AB = ⋅() () is (a) A + B (b) AB (c) A B + (d) AB AB + 51. The reduced form of the Boolean expression AB CAB AC [ ( )] + + is (a) AB (b) AB (c) AB (d) AB AC + 52. Which of the following statements is not correct? (a) X XY X + = (b) X X Y XY + + = ( ) (c) X X Y XY + + = (d) ZX ZXY ZX ZY + = + 53.A system B satisfying IIV is a Boolean Algebra. That is, as a consequence of IIV above, ue shall have V. x+0=x,x.0=0. VI. S=jc,0=1; f=0. VII. x+x=x, x.x=x, x(x+y)=x, x+x.y=x. VIII. x +y.z =(x +y)(x +z). IX. x+(y+z)=(x+y)+z, x(yz)=(xy)z. The proof of x.x=x. Proof. — x.x=x.x+0=x.x+x.x=x(x+x)=x.l =x(I, IV, II, III). 4. x=x. Proof. — x=l.x=(x +x)x=x.x +x.x=x.x +0 =x.x~+x.x=x(x~+x)=x.l=x(III, IV, II I). 5. 0 = 1,7=0. These follow from IV.and 1,2 when x=0 and x=l respectively. 6. If x+y=l and x.y=0, then y=x.The correspondence pl from X | to Yl involved in the above proposition is called the descent of the correspondence p from X to Y in V*). (4) The descent of the composite of correspondences inside V(*) is the composite of their descents: (V o (p)l = Vlo pl. (5) If p is a correspondence inside V* then (p")l = (pl.)". (6) Let Idx be the identity mapping inside V* of a set X e V*). Then (Idź)l = Idx1. (7) Suppose that X, Y, f e V*) are such that [f : X → Y = 1, i.e., f is a mapping from X to Y inside V*).According to the twovalued Boolean algebra x + 1 = 1. [Ans. True') 17. According to the twovalued Boolean algebra (x')' = x. [Ans. True] 18. Truth.table gives the relation between all possible combinations inputs and output. [Ans. True] 19. The twoelement Boolean algebra is called as switching algebra. [Ans. True] 20. The complement of a sum is equal to the both minterm values must be differentiated by power of 2] 6. x = xy = [Ans. X] 7. AB+A+1 [Ans.l] 8. A Prime Implicant Chart in