Hello shinhidora Originally Posted by

**shinhidora** So I'm preparing for my exam on monday and I found this exercise hanging around in my math book, been staring at it for some time now but still no clue where to start...

Simplify the given Boolean formula:

$\displaystyle x' (yzu + z' + u') + (x(y'+u))'$

Edit: not sure if this was Advanced Algebra, so I just put it with Algebra

Using De Morgan's Law twice:

$\displaystyle (x(y'+u))' = x'+(y'+u)'$$\displaystyle =x'+yu'$

So

$\displaystyle x'(yzu+z'+u') + (x(y'+u))'=x'(yzu+z'+u')+x'+yu'$

$\displaystyle =x'+yu'$, since $\displaystyle x'p+x' = x'(p+1)=x'$ where $\displaystyle p = (yzu+z'+u')$

Grandad