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I tried distributing but it just make it worse and lengthy

(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

nope, X*X = 0

I mean X*X = X and X*X'=0

lool

It's different than the other math you are thinking :[

the dot which looks like multiplication is not multiplication. Stands for AND

I mean OR! and the + stands for AND

:(

I guess I'll take the long path

(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

Expand it and then use that identity rule. Did that not cut many things down?

Do you want me to distribute it all? (working on it)

must show all work

are you allowed to use kmap ?

Alright, its going to be tricky, lets see...

WY'X + WY'Z' + W'X'Y + W'X'Z
step1 : group first two terms and last two terms
WY'(X+Z') + W'X'(Y+Z)

How did you get that?

oh wait you are looking at the answer

I am trying to obtain WY'X + WY'Z' + W'X'Y + W'X'Z from (W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)

So the problem I'm simplifying is
(W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)

Ohk.. I thought your started with WY'X + WY'Z' + W'X'Y + W'X'Z
lets start over

How did you applied X ( X + Y ) = X on step 2?

call it U(U+V) = U
U = W+X'
V = Z'

ooooo magic

not really
we're just using the given theorems

(W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)

(W'+Y')(W'+X+Z')(W+X')(W+Y+Z) (brain still loading trying to figure out next step)

(W+X′)(W+Y+Z)
if 8D = X + YZ = ( X + Y ) ( X + Z )
U+BC = (U+B)(U+C)
U= W
B=X'
C=(Y+Z)?

(W′+Y′)(W′+X+Z′)
U+BC = (U+B)(U+C)
U=W'
B=Y'
C=X+Z'
W'+Y'(X+Z')

Yes

finally use theorem 16

W+X′(Y+Z) = (W+X')(Y+Z) right?

since W+X'(Y+Z) is way different and idk how to apply theorem 16 to it :[

How did you applied it? (( X + Y ) ( X' + Z ) = X Z + X' Y)

(U+V)(U'+T) = UT + U'V
U = W
V = X'(Y+Z)
T = Y'(X+Z')

for example, can you explain why below holds ?
X + X' = 1

|dw:1441253021508:dw| my brain went completely blank when I was doing that

no, im unsure

you're confusing between AND gate and OR gate

X + X'
is read as
X or X'
|dw:1441253525157:dw|

ow yes, my bad

it spits out 1 when at least one of its inputs is 1

easy to see that one of the X or X' is always 1

because X, X' are opposites of each other
if X is 0, then X' will be 1
and viceversa

Why is it that X+0=X is not = 1?

X + 0
let X = 0
what does previous expression evaluate to

If X = 0 then we have 0 + 0 which is = 0

so X + 0 = 1 does not always hold