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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)'?

http://www.indiabix.com/digital-electronics/boolean-algebra-and-logic-simplification/discussion-138

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  • \$\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
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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.

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\$\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.

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  • \$\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
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    \$\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
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    \$\begingroup\$ @Carter what ? make what sure ? \$\endgroup\$ – efox29 Feb 2 '15 at 5:53

protected by Nick Alexeev Aug 4 at 21:54

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