Simplify the Boolean function (x cdot y)+[(x+ | vedclass

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(B) Given expression: $(x \cdot y)+[(x+y') \cdot y]'$ Using De Morgan's Law $(a \cdot b)' = a' + b'$: $= (x \cdot y) + [(x+y')' + y']$ Using De Morgan

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$1$

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(B) Given expression: $(x \cdot y)+[(x+y') \cdot y]'$ Using De Morgan's Law $(a \cdot b)' = a' + b'$: $= (x \cdot y) + [(x+y')' + y']$ Using De Morgan's Law $(a+b)' = a' \cdot b'$ and involution law $(y')' = y$: $= (x \cdot y) + [x' \cdot (y')' + y']$ $= (x \cdot y) + [x' \cdot y + y']$ $= x \cdot y + x' \cdot y + y'$ Factor out $y$ from the first two terms: $= y \cdot (x + x') + y'$ Since $x + x' = 1$: $= y \cdot 1 + y'$ $= y + y'$ Since $y + y' = 1$: $= 1$

Simplify the Boolean function $(x \cdot y)+[(x+y') \cdot y]'$

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