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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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How many two-input AND and OR gates are required to realize Y=CD+EF+G ? Ans. Y=CD+EF+G No. of two i/p AND gates=2 No. of two i/p OR gates = 2 One OR gate to OR CD and EF
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Simplify the given expressions using Boolean postulates Y = (A + B)(A‾ + C)(B + C) Ans. Y = (A + B)(A‾ + C)(B + C) = (A A‾ + AC + B A‾ + BC) (B + C) = (AC + B A‾ + BC) (B + C)
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