I'm trying to learning boolean algebra. One of de morgan's law is not understood. how does A'B'C' convert to A'+B'+C'?

As I know,
1. (A'B')' = A+B
2. (AB)'=A'+B'
3. (A+B)'=A'B'
I think it should be (A+B+C)'.

Is this same between A'+B'+C' and (A+B+C)' or A'B'C' and (ABC)'?


  • \$\begingroup\$ Just make a big truth table, once you see it for yourself it will be totally clear. Play around with different statements, you will learn it easily. \$\endgroup\$ – WalyKu Feb 3 '15 at 11:06

In your post, you say:

2. (AB)'=A'+B' 
3. (A+B)'=A'B'
I think it should be (A+B+C)'.  <== I don't know where this fits in

then you provide four combinations:

A'+B'+C' and (A+B+C)' or A'B'C' and (ABC)' 

Looking at each one,

A'+B'+C' = (ABC)'     (flipping 2 around and extending to three terms)
(A+B+C)' = A'B'C'     (from 3, extending to three terms)
A'B'C' = (A+B+C)'     (flipping 3 around and extending to three terms)
(ABC)' = A'+B'+C'     (from 2, extending to three terms)

so the first and last are the same, and the middle two are the same.


\$\bar A \bar B \bar C \$ does not equal \$ \bar A+ \bar B + \bar C \$

If you let A=0, B= 1, and C=1

Then \$\bar A \bar B \bar C \$ = 0 and \$ \bar A+ \bar B + \bar C \$ = 1

It's clear that they are not the same

If you apply deMorgan's for \$\bar A \bar B \bar C \$ you would end up with (A+B+C)'. Which is what you said.

  • \$\begingroup\$ Thanks, so, is this wrong explanation? indiabix.com/digital-electronics/… \$\endgroup\$ – Carter Feb 2 '15 at 1:02
  • \$\begingroup\$ The explanation from privithi is correct. He doesn't make the claim that A'B'C' = A'+B'+C' - which is what this question is asking. \$\endgroup\$ – efox29 Feb 2 '15 at 1:08
  • 4
    \$\begingroup\$ No, the explanation is correct. It's just that the solid bar over ABC means the total expression ABC is inverted, not the individual terms. \$\endgroup\$ – WhatRoughBeast Feb 2 '15 at 1:51
  • \$\begingroup\$ @WhatRoughBeast Whose answer is right? What is different between total inverted and individual divided? \$\endgroup\$ – Carter Feb 2 '15 at 5:50
  • 1
    \$\begingroup\$ @Carter what ? make what sure ? \$\endgroup\$ – efox29 Feb 2 '15 at 5:53

protected by Nick Alexeev Aug 4 at 21:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.