1. ## Simplifying Boolean Algebra

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:

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

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

x´(yuz +1 +2u´+y´´) +z´ ???

3. Originally Posted by Henryt999
x´(yuz +1 +2u´+y´´) +z´ ???
Since it's Boolean (0 & 1) algebra, I think it's quite impossible there's a 2 in the answer...

4. Hi

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

First, $(x(y'+u))'=x'+yu'=x'(y+y')+yu'=x'y+x'y'+yu'$

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

Finally, note that $(yzu)'=z'+u'+y'$. So $yzu+z'+u'+y'=1$

---> $x' (yzu + z' + u') + (x(y'+u))'=x'+x'y+yu'$

5. 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:

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

Edit: not sure if this was Advanced Algebra, so I just put it with Algebra
Using De Morgan's Law twice:
$(x(y'+u))' = x'+(y'+u)'$
$=x'+yu'$
So
$x'(yzu+z'+u') + (x(y'+u))'=x'(yzu+z'+u')+x'+yu'$
$=x'+yu'$, since $x'p+x' = x'(p+1)=x'$ where $p = (yzu+z'+u')$

6. Originally Posted by Moo
Hi

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

First, $(x(y'+u))'=x'+yu'=x'(y+y')+yu'=x'y+x'y'+yu'$

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

Finally, note that $(yzu)'=z'+u'+y'$. So $yzu+z'+u'+y'=1$

---> $x' (yzu + z' + u') + (x(y'+u))'=x'+x'y+yu'$
Thx Moo

But was just wondering...

$x'+x'y+yu' = x' + yu'$ ?

7. Originally Posted by shinhidora
Thx Moo

But was just wondering...

$x'+x'y+yu' = x' + yu'$ ?