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I am asked to simply the following expression

$$F(a,b,c) = c’ab + c’b’ + aba + b’cb + abc + c’b$$

using the Boolean identities and finding $F'(a, b, c)$ using DeMorgan’s law

I have been trying for days but I assume I cannot reduce it properly. your help is tremendously appreciated!

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1 Answer

$b'cb$ is false since it contains both $b$ and $b'$, so we can drop that term.

$c'b'+c'b=c'(b'+b)=c'$ and $c'ab+abc=ab(c'+c)=ab$ and $aba=ab$ and $ab+ab=ab$, so

$$ F(a,b,c)=c'+ab\;. $$

Then

$$ F'(a,b,c)=(c'+ab)'=(c')'(ab)'=c(a'+b')=ca'+cb'\;. $$

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