I have the same question. Someone on stack overflow gave me this answer(i don't know if its correct, because i don't know if we can use the ^:
!(a + b + c) || (b * a ^ c)
where ^ is xor.
you can see my post go to site stackoverflow
search-> Boolean algebra simplification to lowest form
Simplify this function using Boolean algebra laws?
I am stuck on this part of my homework
(x2'x1'x0') + (x2'x1x0) + (x2x1x0')
I know you have to use distributive laws to do this.
Any help is appreciated.
Thank you.
1 answer
1 answer