# Thread: DeMorgan's theorem

1. ## DeMorgan's theorem

Hello,

I need to use the first equation of E to derive the second equation for E. I've been messing around with this for over an hour, and can't seem to find a way to do it. Could someone provide some insight?

Thanks so much,

2. Hello, centenial!

$E \;=\;(A\cdot B\cdot \overline{C}) + (A\cdot \overline{B}\cdot C) + (\overline{A}\cdot B\cdot C)$

$\text{We can also write }E\text{ as a product of sums:}$

. . $E \;=\;\overline{(\overline{A} + \overline{B} + C)\cdot (\overline{A} + B + \overline{C}) \cdot (A + \overline{B} + \overline{C})}$

$\text{To derive this, you need to use DeMorgan's Theorems.}$

It's much easier to work "backwards" . . .

We have: . $E \;=\;\overline{(\overline{A} + \overline{B} + C)\cdot (\overline{A} + B + \overline{C}) \cdot (A + \overline{B} + \overline{C})}$

. . . . . . . . . . $=\;\overline{(\overline A + \overline B + C)} + \overline{(\overline A + B + \overline C)} + \overline{(A + \overline B + \overline C)}$

. . . . . . . . . . $=\; (A\cdot B\cdot\overline C) + (A\cdot\overline B\cdot C) + (\overline A\cdot B\cdot C)$

3. Thanks! That makes sense now.