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 (𝑏+𝑑)β‹…1 previously in the equation?

Also 𝑏+𝑐 into 1 while we have (𝑏+𝑐)β‹…1 previously in the equation?

Picture from that question:

enter image description here


1 Answer 1


They applied the Absorption Law 1 from Boolean Algebra. According to this law, the following is true:

$$x \land (x \lor y) = x$$

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.

(By the way, pictures like this aren't welcomed on this site)


Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.