Prove Boolean Equation: (A'B')'+D'=A+(BD)'
Prove Boolean Equation: (A'B')'+D'=A+(BD)'
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Thank you. I can see that you have now applied De Morgan's the same way I do (I'm still confused as to how you were applying it previously). However, for the line suggesting that we apply the Boolean identity X+X'=1, with X equaling A+B, I would have expected to see (A+B)+(A+B)' being applied somewhere, which I don't. The equation is simply rewritten with no changes (no applications of an identity) at all. Could you please show how this identity was applied so that I can understand it? Thanks, again.
Still not getting how you are applying De Morgan's Law. (A'B')' would be A''+B'', which would simplify to A+B, and (BD)' would be B'+D'. However, you are getting A'+B and B'+D instead. Yet, you say I am right even though we get different expressions. This is very confusing. Are you skipping (not showing) some steps?
According to De Morgan's Law, (A'B')' would be A''+B'', which would simplify to A+B, and (BD)' would be B'+D'. This much I think I'm pretty solid on. However, if I'm correct, it makes the given solution completely incorrect. There are a lot of other parts of the given solution that are very unclear as well. Please try again with better explanations and showing of work. Thank you...