Question Number 96811 by Ar Brandon last updated on 05/Jun/20
$$\mathrm{Prove}\:\mathrm{that}; \\ $$$$\mathrm{a}\backslash\:\mathrm{A}+\mathrm{A}\centerdot\mathrm{B}=\mathrm{A}\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}\backslash\:\left(\mathrm{A}+\mathrm{B}^{} \right)\centerdot\left(\mathrm{A}+\overset{−} {\mathrm{B}}\right)=\mathrm{A} \\ $$$$\mathrm{b}\backslash\:\mathrm{A}\centerdot\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{A}\:\:\:\:\:\:\:\mathrm{d}\backslash\:\mathrm{A}+\overset{−} {\mathrm{A}B}=\mathrm{A}+\mathrm{B} \\ $$
Answered by 1549442205 last updated on 05/Jun/20
$$\mathrm{your}\:\:\mathrm{symbol}\:\mathrm{isn}'\mathrm{t}\:\mathrm{normal}:\mathrm{what}\:\mathrm{does}\:\mathrm{A}.\mathrm{B}\:\mathrm{means}?\mathrm{A}\cap\mathrm{B}?\mathrm{A}+\mathrm{B}=\mathrm{A}\cup\mathrm{B}? \\ $$
Commented by Ar Brandon last updated on 05/Jun/20
It has to do with logic gates.