Question

Boolean algebra
simplify
a'd(b'c)+a'd'(b+c')+(b'+c)(b+c')

Answer 1

The last term can be simplified a lot

Answer 2

you can factor a' from the first 2 terms

Answer 3

and since there is a d' and d it will cancel

Answer 4

Let me type it up.

Answer 5

for last two terms
(b'+c)(b+c')
=b'b+b'c'+bc+cc'
=b'c'+bc

Answer 6

Do you think you can tell me the theorems that you used?

Answer 7

1 or 0 = 1
1 & 1 = 1
1 & 0 = 0
0 or 0 = 0

Answer 8

sorry. i think i got a mistake
i used d instead of d'

Answer 9

D'D=0, as one of them is 1 and other other 0
D'+D=1, "

Answer 10

you can factor and expand brackets an the commutative and associative laws work the same as with normal maths.

Answer 11

oops a mistake T_T

Answer 12

its fine take your time haha, thanks for your help by the way.

Answer 13

isnt bc+c'b' same as XNOR

Answer 14

yeah wouldn't that be the uniting theorem?

Answer 15

nvm thats xy+xy' = x its not a double prime.

Answer 16

any other simplification you see there?

Answer 17

do a karnaugh map instead

Answer 18

It's easier to simplify with it

Answer 19

I haven't really been taught truth tables and kmaps. Do you think you can explain?

Answer 20

@dan815 i think in your 3rd step ,
if you take d' common then in bracket (b+c') should be there instead of (b+c)