# Boolean algebra transformation

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.

• 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. Oct 2 '20 at 23:10

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 $$\overline{A+ B}=\overline{A} \cdot \overline {B}$$: $$A\cdot {\overline {B}}+{\overline {A}}\cdot B \equiv \overline { \overline{A}+B}+\overline {A+\overline {B}} \equiv \\ \overline {(\overline{A}+B) \cdot (A+\overline {B})} \equiv \overline {A\cdot B + \overline{A} \cdot \overline {B}} \equiv \overline{A\cdot B} \cdot \overline{\overline{A} \cdot \overline {B}} \equiv \\ (A+B)\cdot ({\overline {A}}+{\overline {B}})$$