Boolean Algebra help

- anonymous

Boolean Algebra help

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

I tried distributing but it just make it worse and lengthy

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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 )

- anonymous

nope, X*X = 0

- anonymous

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

- anonymous

lool

- anonymous

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

- anonymous

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

- anonymous

I mean OR! and the + stands for AND

- anonymous

:(

- anonymous

I guess I'll take the long path

- anonymous

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

- zzr0ck3r

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

- anonymous

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

- zzr0ck3r

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

- anonymous

must show all work

- ganeshie8

are you allowed to use kmap ?

- anonymous

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

- ganeshie8

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

- anonymous

(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)

- anonymous

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)

- ganeshie8

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)

- anonymous

How did you get that?

- anonymous

oh wait you are looking at the answer

- anonymous

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)

- anonymous

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

- ganeshie8

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

- ganeshie8

\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\)
step1 : rearrange the product
\((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)
step2 : use 10D on first two terms
\((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)

- anonymous

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

- ganeshie8

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

- anonymous

ooooo magic

- ganeshie8

not really
we're just using the given theorems

- anonymous

(W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)

- anonymous

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

- ganeshie8

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

- anonymous

(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)?

- anonymous

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

- ganeshie8

Yes

- ganeshie8

finally use theorem 16

- anonymous

W+X′(Y+Z) = (W+X')(Y+Z) right?

- anonymous

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

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- ganeshie8

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

- anonymous

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

- anonymous

no, im unsure

- ganeshie8

you're confusing between AND gate and OR gate

- ganeshie8

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

- anonymous

ow yes, my bad

- ganeshie8

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

- ganeshie8

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

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- anonymous

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

- ganeshie8

so X + 0 = 1 does not always hold

Looking for something else?

Not the answer you are looking for? Search for more explanations.