Quote (shenk @ Mar 9 2014 06:34pm)
thank you
can you tell me what i am comparing x to in the last 2 columns?
ok i get the last column but not the one before it
am i comparing X|!(Y&Z) to the x column?
if so then i get it now
You are ORing X and the result of !(Y&Z) together. Whats not to understand?
So you have three input variables columns depicting each possible state for the three variables. From there you are simply answering for each sub logic in the equation.
X|!(Y&Z)
where X = 1, Y = 0, Z = 1 would be
(0 & 1) = 0
!(0) = 1
1 | 1 = 1
So X|!(Y&Z) where X = 1, Y = 0, Z = 1 would equate to 1.
This post was edited by AbDuCt on Mar 9 2014 05:05pm