Question

let's see:
proof x y' + x' z +y z' =x' y + x z' + y' z

Answer 1

btw, it is boolean

Answer 2

boolean ?

Answer 3

Thanks for the medal :) , obviously it can be proven with a truth table - but that's kinda missing the point probably. I'm reviewing boolean algebra material ( it's been years since I used it) - and will play around with it more :)

Answer 4

ab+a'c+bc <-> ab+a'c
consensus theorem
a b + a' c + (a+a')bc
a b + a' c + abc + a'bc
a b + abc + a'bc+a'c
ab(1+c) + a'c(b+1)
ab+ a'c

Using consensus theorem above we can expand our equation
x y' + x' z +y z' =x' y + x z' + y' z

left side ->we get additional term from x y' + x' z +y z'
y'z+ x'y + x z'
put together with original equation
x y' + x' z +y z'+y'z+ x'y + x z'

----------------------------------------------------
right side-> additional term from x' y + x z' + y' z
yz'+x'z+xy'

put together with original equation
x' y + x z' + y' z+yz'+x'z+xy'



Now compare L and R
x y' + x' z +y z'+y'z+ x'y + x z' =x' y + x z' + y' z+yz'+x'z+xy'

Answer 5

ok - got it..
You applied the consensus theory three times on the left side to get three extra terms.
Did the same to the right side.
And the extended left and right sides became the same.

Very nice.
Do you have a link to a cheat-sheet with more theorems like the consensus theorem ?

Answer 6

Demorgan's law is neat too
(a+b+c)'=a' b' c'

(a b c)'=(a'+b'+c')

Answer 7

yeah, I did remember this one :)

Answer 8

This is consensus too
(a+b)(b+c)(a'+c)=(a+b)(a'+c)