How does $(xy' + x'y)'$ simplify to $(xy + x'y')$

$\begingroup$

How does

$$(xy' + x'y)'$$

simplify to

$$(xy + x'y')$$

$\endgroup$ 2

1 Answer

$\begingroup$

By de Morgan laws : $$(xy'+x'y)'=(x'+y)(x+y')=x'x+x'y'+yx+yy'=xy+x'y'$$ because $xx'=yy'=0$.

(BTW, I suppose $x'$ means $\overline x$, or $\neg x$...)

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy