# What boolean algebra simplification is applied in this Shannon's expansion?

Found this question about Shannon's expansion. While I am trying to follow its logic, found one super convenience simplification used. Can we do this in general while dealing with boolean algebra? or only in Shannon's expansion?

Can we just turn b+d into 1 while we have (π+π)β1 previously in the equation?

Also π+π into 1 while we have (π+π)β1 previously in the equation?

Picture from that question:

• Oct 13, 2021 at 0:07
• Thank you for the link, but at the same time. wonder if you read the question. Oct 13, 2021 at 0:34

$$x \land (x \lor y) = x$$
Sometimes the operator "$$\land$$" is replaced by multiplication, and also the operator "$$\lor$$" - by addition. That's why:
$$(b+d) \cdot 1 \cdot (b+c+d+e+f)=(b+d) \cdot ((b+d)+c+e+f)=b+d$$ $$1 \cdot (b+c) \cdot (b+c+d+e+f)=(b+c) \cdot ((b+c)+d+e+f)=b+c$$
Yes, appearance of $$1$$ in $$(1+c+e+f)$$ and $$(1+d+e+f)$$ terms looks strange to me as well - may be, they teach this kind of simplifications this way, it's difficult to say without looking at your textbook.
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