Question

Find the dual of the following Boolean expression.
xy+xy
There are two lines over the second x and y.

Answer 1

Not sure how to write this problem out.

Answer 2

\(\large x\bar{y} + \bar{x} y\)

like this ?

Answer 3

nope it is over the last x and y

Answer 4

\(\large xy + \bar{x} \bar{y}\)

Answer 5

yes

Answer 6

two steps :
1. swap AND and OR operations
2. simplify

Answer 7

\(\large xy + \bar{x} \bar{y}\)

1. swap AND and OR operations

\(\large (x+y) ( \bar{x}+ \bar{y})\)

Answer 8

multiply and simplify

Answer 9

okay

Answer 10

\(\large xy + \bar{x} \bar{y}\)

1. swap AND and OR operations

\(\large (x+y) ( \bar{x}+ \bar{y})\)

\(\large (x\bar{x}+ y\bar{x} + x \bar{y} + y \bar{y})\)

Answer 11

use \(x\bar{x} = 0\). il leave the last step for you

Answer 12

With *duality* you merely replace \(+\) with \(\cdot\) as well as \(x\) with \(\bar{x}\):$$xy+xy\to(\bar{x}+\bar{y})(x+y)$$As @ganeshie8 pointed out, we can distribute:$$x\bar{x}+\bar{x}y+x\bar{y}+y\bar{y}$$Logically, we know \(\bar{x}x=0\) and \(y\bar{y}=0\)$$\bar{x}y+x\bar{y}$$

Answer 13

last step is what as I got this far.

Answer 14

whats your last step

Answer 15

not sure still a little lost

Answer 16

its okay... digital logic seems scary at first... but it turns out easy to learn in no time..

Answer 17

you familiar wid AND and OR gates, dont you ?

Answer 18

a little bit

Answer 19

then you may knw \(x\bar{x} = 0\)

Answer 20

okay

Answer 21

AND of \(x\) , \(\bar{x}\) is 0

Answer 22

when you AND a thing, and its negation, one of them is 0 for sure, making the result 0

Answer 23

cuz, AND gate must have all 1s at its input, for it to give 1 as output

Answer 24

Okay