I've got an equation down to $F=BA+BC +\bar{A}\bar{C}$ and according to Wolfram Alpha it can simplified to $(\bar{A} + \bar{C}) + B$. What's the next steps? I tried using de Morgan's law and not sure it helped.
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Try to restrict to one question power post while showing your effort – Shailesh Sep 14 '15 at 06:59
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I think you misread the result. – Stefan Perko Sep 14 '15 at 07:20
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@StefanPerko so it can't be simplified any further? – leftandright Sep 14 '15 at 07:26
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Who's responsible for the -1? – leftandright Sep 14 '15 at 07:34
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This is a derivation for the actual result:
$BA + BC + \bar{A}\bar{C} = (B(A+C))+\overline{A+C} = (B+\overline{A+C})((A+C)+\overline{A+C}) = \overline{A+C}+B$
I used:
- Distributivity of "$\cdot$ over $+$" and DeMorgan (for "NOR")
- Distributivity of "$+$ over $\cdot$"
- $P + \bar{P} = 1$ and $P\cdot 1 = P$
and some Commutativity and Associativity.
Stefan Perko
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@leftandright No, it is not. You know the deMorgan laws, don't you? It's $\overline{A+C} = \bar{A}\bar{C}$ and $\overline{AC} = \bar{A}+\bar{C}$. Using an example: "It is not raining or snowing" (meaning there is no rain and there is no snow) has a different meaning than "It is not raining or it is not snowing." – Stefan Perko Sep 14 '15 at 07:40