$\begingroup$Hi I am new to the board. Taking a computer architecture course and I am having trouble understanding further simplification on a question I got on a previous quiz. When I type in the expression after setting up my K map and simplifying I come to this:
(A AND B) OR (NOT A AND NOT B).
Through the rules of simplification A+ NOT A=1. Does this then mean that AB +A'B'=1? and if so how is that represented as a logic gate?
Thanks.
$\endgroup$32 Answers
$\begingroup$Let's have the truth table of AB + A'B'.
A | B | AB + A'B' --+---+-------------- 0 | 0 | 0.0 + 1.1 = 1 0 | 1 | 0.1 + 1.0 = 0 The assertion that AB + A'B' =1 is definitely not true. Now the source of confusion seems to be the following.
If A + A' = 1 then should AB + A'B' is also equal to 1. NO. The complement of AB is (AB)' and not A'B'.
Use de-Morgan theorem on (AB)' and you get the following,
(AB)' = A' + B' BOOLEAN GATES REPRESENTING (AB)'
+----------+ +--------+ A ----| | | | | & |-----| Inv |---- (AB)' B ----| | | | +----------+ +--------+ A'B' would be.
A---o|---| | & |----A'B' B---o|---| The truth tables for (AB)' is:
(AB)' FF=T, FT=T, TF=T, TT=F. Where as the truth table for A'B' is
A'B' FF=T, FT=F, TF=F, TT=F. 
