Boolean Expression simplification help

Question

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.

Answer 1

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 ----|          |     |        |
         +----------+     +--------+

Answer 2

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.