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I read online that $${\displaystyle (A\cdot {\overline {B}})+({\overline {A}}\cdot B)\equiv (A+B)\cdot ({\overline {A}}+{\overline {B}})}$$

and I can verify this, but I'm not quite sure how to basically take the left-hand side and transform it into the right-hand side. Any help would be appreciated, thank you.

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  • $\begingroup$ Note that $A \bar{A}$ is always false so you can always 'or' it to something. So $A \bar{A} + B \bar{B}$ can be added (well, 'ored') without changing the value. $\endgroup$
    – copper.hat
    Oct 2 '20 at 23:10
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  1. As right hand can be transformed in way: $$(A+B)\cdot ({\overline {A}}+{\overline {B}}) \equiv \\ A \cdot \overline {A}+A \cdot\overline {B} + B \cdot \overline {A} + B \cdot \overline {B } \equiv \\ A\cdot {\overline {B}}+{\overline {A}}\cdot B$$ and knowing, that equivalence works in both direction, you can now write this equivalences in opposite direction to reach desired.

  2. For second way let's use $\overline{A\cdot B}=\overline{A} + \overline {B}$ and $\overline{A+ B}=\overline{A} \cdot \overline {B}$: $$A\cdot {\overline {B}}+{\overline {A}}\cdot B \equiv \overline { \overline{A}+B}+\overline {A+\overline {B}} \equiv \\ \overline {(\overline{A}+B) \cdot (A+\overline {B})} \equiv \overline {A\cdot B + \overline{A} \cdot \overline {B}} \equiv \overline{A\cdot B} \cdot \overline{\overline{A} \cdot \overline {B}} \equiv \\ (A+B)\cdot ({\overline {A}}+{\overline {B}}) $$

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  • $\begingroup$ so there is no other way to transform directly the LHS to the RHS? $\endgroup$
    – Evan
    Oct 3 '20 at 18:41
  • $\begingroup$ @Evan. Why no? There are so many ways, as many fantasy is in head. Added second one to answer (only do not ask me about third way). $\endgroup$
    – zkutch
    Oct 4 '20 at 1:28
  • $\begingroup$ awesome thank you so much!! $\endgroup$
    – Evan
    Oct 4 '20 at 4:30

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