# De Morgan's laws

(Redirected from DeMorgan's Laws)

De_Morgan's laws (or De_Morgan's theorems) are used to simplify the Boolean expressions.

There are two theorems:

1. The complement of two or more AND variables is equal to the OR of the complements of each variable.
${\displaystyle {\overline {xy}}={\bar {x}}+{\bar {y}}}$
2. The complement of two or more OR variables is equal to the AND of the complements of each variable.
${\displaystyle {\overline {x+y}}={\bar {x}}{\bar {y}}}$

Note that this theorems are true in both direction

## Examples

• ${\displaystyle {\overline {xyz}}={\bar {x}}+{\bar {y}}+{\bar {z}}}$
• ${\displaystyle {\overline {x+y+z}}={\bar {x}}{\bar {y}}{\bar {z}}}$
• Note that x can be a combination of other variables.

lets say ${\displaystyle x=(A+B)}$ and ${\displaystyle y=(C+D)}$ so:

${\displaystyle {\overline {xy}}={\bar {x}}+{\bar {y}}}$

will be:

${\displaystyle {\overline {(A+B)(C+D)}}={\overline {(A+B)}}+{\overline {(C+D)}}={\bar {A}}{\bar {B}}+{\bar {C}}{\bar {D}}}$
• In case we have more than one livile bar (or complement or negate) start with the top bar first
${\displaystyle {\overline {{\overline {(A+B)}}+{\overline {(C+D)}}}}={\overline {\overline {(A+B)}}}.{\overline {\overline {(C+D)}}}=(A+B)(C+D)}$

Note that ${\displaystyle {\overline {\overline {A}}}=A}$ Boolean rule.

• ${\displaystyle {\overline {{\overline {ABC}}+D+E}}={\overline {\overline {ABC}}}{\bar {D}}{\bar {E}}=ABC{\bar {D}}{\bar {E}}}$
• ${\displaystyle {\overline {(A+B){\bar {C}}+D+{\bar {E}}}}=({\overline {(A+B){\bar {C}})}}{\bar {D}}{\overline {\overline {E}}}=({\overline {(A+B)}}+{\overline {\overline {C}}}){\bar {D}}E=({\bar {A}}{\bar {B}}+C){\bar {D}}E}$