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I've managed to simplify $\bar{a}\bar{b}(\bar{a} + b)(\bar{b} + b)$ to $\bar{a} +\bar{a}\bar{b}$

Where $\bar{a}$ = $a$ complement I simplified the original expression with Huntingtons Postulates, Idempotent laws etc

I've checked they are equivalent by truth table. Can this be simplified even further?

Dravid
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  • What is the topic? What is $a$. What is $\overline{a}$? What did you try so far? Please expand the question. – Kolja Dec 16 '22 at 16:07
  • @Kolja Sorry if the question is vague, I will edit it now. The topic is Boolean algebra/functions – Dravid Dec 16 '22 at 16:09

1 Answers1

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Recall that $xx = x$, $\bar{x} + x = 1$, and $\bar{x}x = 0$ for all $x$.

\begin{align} \bar{a}\bar{b}(\bar{a} + b)(\bar{b} + b) &= \bar{a}\bar{b}(\bar{a}+b) \\ &= \bar{a}\bar{b}\bar{a} + \bar{a}\bar{b}b \\ &= \bar{a}\bar{a}\bar{b} \\ &= \bar{a}\bar{b}. \end{align}

Line 1: $\bar{x}+x = 1$,
Line 2: $x(y+z) = xy + xz$,
Line 3: $\bar{x}x=0$,
Line 4: $xx = x$.

Kolja
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    I really confused myself by implementing the complement law incorrectly... Thanks for clearing that up! – Dravid Dec 16 '22 at 16:27