Is there any easy way to solve boolean function and find Literals?😅

Question

i couldn't do this

proove,by dual
(x+y)(x'+z)(y+z)=(x+y)(x'+z)

should applying complement on LHS?

Answer 1

you mean LHS, RHS?
L.H.S.
left-hand side
R.H.S.

right-hand side😅😅

Answer 2

Zeke Williams i meant,boolean postulates

i dont know in-depth of it, but we are taught this in school to find literals
or
to simplify make function short in length

Answer 3

yeah, yours a good idea, But I just solved it, sorry for trouble its Been Hours and i just solved it now

taking complement of whole function LHS.
then RHS.
demorgan law

x'y'+xz'+y'z'=x'y'+ xz'

prooving the above statement now by postulates,
multiply
(x+x') to y'z'
simplify and tada proof🤓

Answer 4

(x+y)(x'+z)(y+z)=(x+y)(x'+z)

LHS:
(x+y)(x'+z)(y+z)

(xx'+xz+x'y+yz)(y+z)

(0+xz+x'y+yz)(y+z)
xyz+xz+x'y+x'yz+yz+yz

xz(y+1)+x'y+yz(x'+1+1)

xz+x'y+yz==LHS

RHS:
(x+y)(x'+z)
xx'+x'y+xz+yz
0+x'y+xz+yz

xz+x'y+yz====RHS

LHS= RHS

PROVED

Answer 5

Please explain more. I'm not sure why this algebra expression has anything to do with Boolean logic. What exactly are you solving?

Answer 6

Alright then! I have no idea what that is about. I've been through 2 years of calculus, so I thought I would know, but I'm glad you solved it :)

Answer 7

:)

Answer 8

Sid haha yeah

Answer 9

Edited

Answer 10

sorry edited again,

Answer 11

But which postulates, and proove what? That's just an expression.

EDIT: By dual what?

Answer 12

Silly

Answer 13

You mean like:
xx' + xz + x'y (y + z) = xx' + xz + x'z
y + z = 1

Answer 14

No caps please internet etiquette

Answer 15

can't understand.... actually I never visited a coding school....
I am totally new in programming...(sorry for bad english)