I have some doubts after learning the principle of the duality of boolean algebra, which defines the dual of an expression is obtained by replacing AND with OR and OR with AND, constant 1s by 0s, and 0s by 1s.
However I encountered a problem with understanding it, which is supposing there is given that a boolean expression A + B = 1 is true, whether or not the dual AB = 0 is also true.
As I obtained a truth table of A + B, the given expression is true when A and B are both 1s (1 + 1 = 1). But I found that the dual may be wrong (1 . 1 = 0)
I would wonder if my concept on the principle is correct or not and the statement is false or my understanding is wrong such that the statement is true actually.
Thanks a lot!!
Best Answer
I think you misunderstood the concept. We replace all the variables and constants in the expression by NOT of them. Thus dual of
A + B = 1is(~A)(~B) = 0which have the same truthtable:- $$A=0,B=0 \implies A+B=1 \space false, (\tilde{}A)(\tilde{}B)=0\space false$$ $$A=1,B=0 \implies A+B=1 \space true, (\tilde{}A)(\tilde{}B)=0\space true$$ $$A=0,B=1 \implies A+B=1 \space true, (\tilde{}A)(\tilde{}B)=0\space true$$ $$A=1,B=1 \implies A+B=1 \space true, (\tilde{}A)(\tilde{}B)=0\space true$$Duality in Boolean Algebra ia an application of more general theory in order theorem : Duality Law