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Question Number 96811 by Ar Brandon last updated on 05/Jun/20

$$\mathrm{Prove}\mathrm{that};$$$$\mathrm{a}\mathrm{\setminus }\mathrm{A}+\mathrm{A}\cdot \mathrm{B}=\mathrm{A}\mathrm{c}\mathrm{\setminus }(\mathrm{A}+{\mathrm{B}}^{})\cdot (\mathrm{A}+\stackrel{-}{\mathrm{B}})=\mathrm{A}$$$$\mathrm{b}\mathrm{\setminus }\mathrm{A}\cdot (\mathrm{A}+\mathrm{B})=\mathrm{A}\mathrm{d}\mathrm{\setminus }\mathrm{A}+\stackrel{-}{\mathrm{A}B}=\mathrm{A}+\mathrm{B}$$

Answered by 1549442205 last updated on 05/Jun/20

$$\mathrm{your}\mathrm{symbol}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{normal}:\mathrm{what}\mathrm{does}\mathrm{A}.\mathrm{B}\mathrm{means}?\mathrm{A}\cap \mathrm{B}?\mathrm{A}+\mathrm{B}=\mathrm{A}\cup \mathrm{B}?$$

Commented by Ar Brandon last updated on 05/Jun/20

It has to do with logic gates.

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