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Simplify the Boolean expression F = C(B + C)(A + B + C).
Ans. Simplification of the given Boolean Expression F = C (B +C) (A+B+C) given asF = C (B+C) (A+B+C)= CB + CC [(A+B+C)]= CB + C [(A+B+C)] (Q CC = C)= CBA + CBB + CBC + CA + CB + CC= ABC + CB + CB + CA + CB + CC (Q CBB =CB & CBC = CB)= ABC + CB + CA + C (Q CB+CB+CB = CB; CC = C)= ABC + BC + C (1+A)= ABC + BC + C (Q 1+A = 1)= ABC + C (1+B)= ABC + C (Q 1+B = 1)= C (1+AB) = C {Q (1+AB)=1}
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Subtraction 11011-11001 using 2's complement. Ans. 11011 - 11001 = A - B 2's complement of B = 00111 1 1 0 1 1 + 0 0 1 1 1 1 0 0 0 1 0 Ignore carry to get answer as 00010 = 2.
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