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Implement the following function using 8 to 1 multiplexerY(A, B, C, D) = ∑(0,1,2,5,9,11,13,15)
Ans. We will obtain three variables B,C and D at selection lines and also A as input. Here eight inputs are given and they can be 0,1, A or A' depending upon the Boolean function.
I0
I1
I2
I3
I4
I5
I6
I7
A'
0 1 2 3 4 5 6 7
8 9 10 11 12 13 14 15
A
1
0
Then, the realization is:
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Q. Prove using Boolean Algebra 1. AB + AC + BC' = AC + BC' 2. (A+B+C) (A+B'+C') (A+B+C') (A+B'+C)=A 3. (A+B) (A'+B'+C) + AB = A+B 4. A'C + A'B + AB'C + BC = C + A'B
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