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I'm currently practicing on Boolean algebra and theorems, recently I found a question that asks to simplify the Boolean expression ABC+A'B'C'. During trying to simplify this when I was trying to construct a KMAP

A/BC 00 01 11 10
0 1 0 0 0
1 0 0 1 0

I realized that I can instead use A XNOR B AND B XNOR C and so I attempted to figure out how to transform it and this is my try:

$$ABC + \overline A \overline B \overline C$$ $$\overline {\overline{(ABC)} \overline{(\overline A \overline B \overline C)}}$$ $$\overline{(\overline A + \overline B + \overline C)(A + B + C)}$$ $$\overline{\overline AB + \overline AC + \overline BA + \overline BC + \overline CA +\overline CB}$$

after this i can group it up and make the XORS but im stuck and i cannot make the 2 XNOR

tritop21
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1 Answers1

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These maybe more obvious by K-map:

$$\begin{align*} \overline AB + \overline AC + \overline BC &= \overline AB + \overline A\left(B+\overline B\right)C + \overline BC\\ &= \overline AB(1+C) + \overline BC\left(\overline A+1\right)\\ &= \overline AB + \overline BC \end{align*}$$

Similarly,

$$\overline BA + \overline CA + \overline CB = \overline BA + \overline CB$$

Then overall,

$$\def\xnor{\mathbin{\text{XNOR}}} \def\xor{\mathbin{\text{XOR}}} \begin{align*} ABC + \overline A \ \overline B \ \overline C = \cdots &= \overline{\overline A B + \overline AC + \overline BA + \overline BC + \overline CA + \overline CB}\\ &= \overline{\overline A B + \overline B A + \overline BC + \overline C B}\\ &= \left(\overline{\overline A B + \overline B A}\right) \left(\overline{\overline BC + \overline C B}\right)\\ &= \left(\overline{A\xor B}\right) \left(\overline{B\xor C}\right)\\ &= (A\xnor B)(B\xnor C)\\ \end{align*}$$

peterwhy
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    ahh, I never thought of expanding on 1 to progress with the equation, that really opened my eyes thankyou very much! – tritop21 Dec 20 '23 at 17:27