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Simplify-the-following-expression-using-the-laws-of-Boolean-algebra-x-y-y-z-x-z-




Question Number 1047 by 112358 last updated on 23/May/15
Simplify the following  expression using the laws of   Boolean algebra:  (x∧∽y)∨(∽y∧∽z)∨(∽x∧∽z) .
$${Simplify}\:{the}\:{following} \\ $$$${expression}\:{using}\:{the}\:{laws}\:{of}\: \\ $$$${Boolean}\:{algebra}: \\ $$$$\left({x}\wedge\backsim{y}\right)\vee\left(\backsim{y}\wedge\backsim{z}\right)\vee\left(\backsim{x}\wedge\backsim{z}\right)\:. \\ $$$$ \\ $$
Commented by prakash jain last updated on 24/May/15
∽x∧∽z=∽(x∨z)  ∽y∧∽z=∽(y∨z)  (∽(y∨z))∨(∽(x∨z))=∽((y∨z)∧(x∨z))  =∽((y∧x)∨z)
$$\backsim{x}\wedge\backsim{z}=\backsim\left({x}\vee{z}\right) \\ $$$$\backsim{y}\wedge\backsim{z}=\backsim\left({y}\vee{z}\right) \\ $$$$\left(\backsim\left({y}\vee{z}\right)\right)\vee\left(\backsim\left({x}\vee{z}\right)\right)=\backsim\left(\left({y}\vee{z}\right)\wedge\left({x}\vee{z}\right)\right) \\ $$$$=\backsim\left(\left({y}\wedge{x}\right)\vee{z}\right) \\ $$$$ \\ $$

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