Question

@rational

Answer 1

DF'G + DE'F' + DE'G' = DF'G + DE'G'

Answer 2

is this logically equivalent?

Answer 3

Maybe or may be not.

what that got to do with the original question :
DE'(F' + G') + DE'(F' + E)

?

Answer 4

you have left me in confused state in previous thread and started a new question asking about a new expression :/

Answer 5

DE'F' + DE'G' = DE'(F'+G') + DE'(F'+E)

Answer 6

that's how they're related

Answer 7

the expression i added was an expression i thought would not matter in simplifying since it is in the answer however i may have been wrong

Answer 8

we cannot add terms except for the identity terms

Answer 9

0+x = x

so you can add 0 if you want

Answer 10

you mean in the sense of a consensus theorem?

Answer 11

like, EE' because it evaluates to 0

Answer 12

but adding DF'G is a crime

Answer 13

yes i know It was part of the original equation

Answer 14

Think of it as me giving you half the equation to try to simplify HALF of the problem

Answer 15

what was the original question ? please provide full details, we're already wasting lots of time without any progress :/

Answer 16

hold on i'm creating a truth table atm

Answer 17

they're logically equivalent

Answer 18

i can attach the file if you want to see

Answer 19

so you can simplify this to DE'F' + DE'G'

Answer 20

so we need to go from DF'G + DE'F' + DE'G' to DE'F' + DE'G'

Answer 21

DF'G + D(E+F)' + D(E+G)'

Answer 22

I don't get what the original question is yet

Answer 23

The question is Simplify DF'G + DE'F' + DE'G'

Answer 24

then what happened to the previous question ?

Answer 25

forget about it

Answer 26

just forget it ?

Answer 27

yes

Answer 28

DF'G + DE'F' + DE'G' = DF'G+ D(E+F)' + D(E+G)'

Answer 29

D( F'G+(E+F)' + (E+G)' )