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My paper is due tomorrow and there is only the last exercise left for me to do. However, I don't have any sufficient notes or examples on how to simplify it. Any help would be appreciated!

A'B'C' + A'B'C + ABC'

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  • $\begingroup$ Hints: $XY \lor XY' = X(Y \lor Y') = X\land 1 = X$ $\endgroup$ Dec 3 '13 at 19:30
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We use the distributive law (DL), and the identities $$P + P' = 1\tag{1}$$ $$ 1\cdot P = P\cdot 1 = P\tag{2}$$


$$\begin{align} \color{blue}{\bf A'B'}C' + \color{blue}{\bf A'B'}C + ABC' &= \color{blue}{\bf A'B'}(C + C') + ABC' \tag{DL} \\ \\ & = A'B'(1) + ABC'\tag{1} \\ \\ & = A'B' + ABC' \tag{2}\end{align}$$

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  • $\begingroup$ Nice color coding +1 $\endgroup$
    – Amzoti
    Dec 4 '13 at 2:44
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$\bar{A}\bar{B}\bar{C}+\bar{A}\bar{B}{C}+{A}{B}\bar{C}=(\bar{A}\bar{B}\bar{C}+\bar{A}\bar{B}{C})+{A}{B}\bar{C}=\bar{A}\bar{B}(\bar{C}+C)+{A}{B}\bar{C}=\bar{A}\bar{B}+{A}{B}\bar{C}$

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