# DeMorgans Law

#### Xander

Joined Dec 13, 2005
6
i need to apply DeMorgans law onto this boolean algebra.

its A.B with 2 bars. one bar over the b and another bar over both of them.

i think it ends up in A + B with both nots but i cant prove it. anyone like to help? and take me through it please?

#### shortcircuit85

Joined Dec 13, 2005
1
Originally posted by Xander@Dec 13 2005, 03:33 PM
i need to apply DeMorgans law onto this boolean algebra.

its A.B with 2 bars. one bar over the b and another bar over both of them.

i think it ends up in A + B with both nots but i cant prove it. anyone like to help? and take me through it please?
[post=12432]Quoted post[/post]​
"bar A*B bar" or "/(A/B)" or "not A B bar" are different ways to describe the function you are talking about.

Applying demorgans is simple. Just remember this. Change the bar, change the sign.

In other words, if you change (break or join) a bar, you need to change the sign from + or * to it's opposite. Keep in mind "A*B" is the same thing as "AB"

For example. /(AB) = /A + /B

Your problem just requires a simple change. You need to recognize two bars = no bars. i.e. //B = B

/(A/B) = /A + //B = /A + B