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I want to write the Product of Sums from of the boolean function $$F(A, B, C) = (A*B) + (A'*A) + (B'*C)$$

I can cancel out $(A'*A)$ and then I'm wondering how I can apply De Morgans law on $$(AB)\cup (B'C)$$

Is it even valid? I just need to get it into the form $(a+b)*(c+d)$ i think

EDIT: My attempt:

$(A\cap B)\cup(B'\cap C) = ((A\cap B)\cup B')\cap ((A\cap B)\cup C) = (A\cup B')\cap (A\cup C) \cap (C\cup B) $

user832075
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1 Answers1

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There maybe a shorter path to reach it, but here it goes one...

\begin{align} AB + B'C &= (AB + B')(AB + C)\\ &= (A + B')(B + B')(A + C)(B + C)\\ &= (A + B')(A + C)(B + C)\\ &= (A + B')(B + C)A + (A + B')(B + C)C\\ &= A(B + C) + (A + B')C\\ &= AB + AC + B'C\\ &= (A + B')(B + C). \end{align}

If you have trouble justifying some passage, let me know.

Another way to go see that $AB+B'C=AB+AC+B'C$ (which takes most of the steps above), is to show that $(AB+B'C)AC = AC$ (essentially use distributivity), and use that in Boolean algebra (or any lattice), if $XY=Y$, then $X+Y=X+XY=X$.

amrsa
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  • I got something very similar: $(A+B')(A+C)(C+B)$. Is this equivalent to yours? Edit: its clear that its not – user832075 Aug 17 '24 at 15:29
  • PLease look at my edit, can you see where I went wrong? – user832075 Aug 17 '24 at 15:29
  • @user832075 Aside the notational inconsistency of going from $*$ to $\cap$, etc, everything is ok. It just can be simplified e bit more, getting the expression in my answer. In fact, you can notice that that is the expression I have in the third line of the aligned environment. – amrsa Aug 17 '24 at 15:35
  • Ah, ok, just noticed the first comment, to which the answer is YES, which I referred to in my previous comment. However, since I have a shorter expression, mine one is preferable ($(A+C)$ in the expression is redundant, like $X1$, which is just $X$, just not obvious). – amrsa Aug 17 '24 at 15:41