Question

Boolean Algebra help

Answer 1

How can I simplify this to get WY'X + WY'Z' + W'X'Y + W'X'Z as an answer? \[(W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\]

Answer 2

I tried distributing but it just make it worse and lengthy

Answer 3

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

Answer 4

Yes, there are some theorems here that can help simplify it faster but I am unsure as to how to proceed with this problem :( (
http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf )

Answer 5

nope, X*X = 0

Answer 6

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

Answer 7

lool

Answer 8

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

Answer 9

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

Answer 10

I mean OR! and the + stands for AND

Answer 11

:(

Answer 12

I guess I'll take the long path

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

I mean I would wolfram...I don't think your teacher will mind. At this level we know how to distribute.

Answer 17

must show all work

Answer 18

are you allowed to use kmap ?

Answer 19

Professor hasn't taught it yet, only theorems
http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf

Answer 20

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

Answer 21

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

Answer 22

W'(WW')=0
(WW'X+WW'Z'+WW'Y+WXY'+WY'Z'+W'W'X+W'X'X+W'X'Z'+W'X'Y'+X'XY+X'Y'Z'+W'W'Z+W'XZ+W'Z'Z+W'Y'Z+XY'Z)(W+X')(W+Y+Z)

Answer 23

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)

Answer 24

How did you get that?

Answer 25

oh wait you are looking at the answer

Answer 26

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)

Answer 27

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

Answer 28

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

lets start over

Answer 29

\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\)

step1 : rearrange the product
\((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

step2 : use 10D on first two terms
\((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

Answer 30

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

Answer 31

call it U(U+V) = U

U = W+X'
V = Z'

Answer 32

ooooo magic

Answer 33

not really
we're just using the given theorems

Answer 34

(W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)

Answer 35

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

Answer 36

\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\)

step1 : rearrange the product
\((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

step2 : use 10D on first two terms
\((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

step3 : rearrange the product
\((W+X')(W+Y+Z)(W'+Y')(W'+X+Z')\)

step4 : use 8D in reverse on first two terms
\([W+X'(Y+Z)](W'+Y')(W'+X+Z')\)

step5 : use 8D in reverse on last two terms
\([W+X'(Y+Z)][W'+Y'(X+Z')]\)

Answer 37

(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)?

Answer 38

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

Answer 39

Yes

Answer 40

finally use theorem 16

Answer 41

W+X′(Y+Z) = (W+X')(Y+Z) right?

Answer 42

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

Answer 43

\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\)

step1 : rearrange the product
\((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

step2 : use 10D on first two terms
\((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

step3 : rearrange the product
\((W+X')(W+Y+Z)(W'+Y')(W'+X+Z')\)

step4 : use 8D in reverse on first two terms
\([W+X'(Y+Z)](W'+Y')(W'+X+Z')\)

step5 : use 8D in reverse on last two terms
\([W+X'(Y+Z)][W'+Y'(X+Z')]\)

step6 : use 16
\(WY'(X+Z') + W'X'(Y+Z)\)

Answer 44

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

Answer 45

(U+V)(U'+T) = UT + U'V

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

Answer 46

What's your thought process when figuring out what theorem to use? I'm having a hard time figuring out what theorem to apply.

Answer 47

you need to understand well why those theorems work
and convince everything in terms of conjunctions and disjunctions

Answer 48

for example, can you explain why below holds ?

X + X' = 1

Answer 49

my brain went completely blank when I was doing that