anonymous
  • anonymous
Boolean Algebra help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
I tried distributing but it just make it worse and lengthy
anonymous
  • anonymous
(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

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More answers

anonymous
  • 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
  • anonymous
nope, X*X = 0
anonymous
  • anonymous
I mean X*X = X and X*X'=0
anonymous
  • anonymous
lool
anonymous
  • anonymous
It's different than the other math you are thinking :[
anonymous
  • anonymous
the dot which looks like multiplication is not multiplication. Stands for AND
anonymous
  • anonymous
I mean OR! and the + stands for AND
anonymous
  • anonymous
:(
anonymous
  • anonymous
I guess I'll take the long path
anonymous
  • anonymous
(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)
zzr0ck3r
  • zzr0ck3r
Expand it and then use that identity rule. Did that not cut many things down?
anonymous
  • anonymous
Do you want me to distribute it all? (working on it)
zzr0ck3r
  • zzr0ck3r
I mean I would wolfram...I don't think your teacher will mind. At this level we know how to distribute.
anonymous
  • anonymous
must show all work
ganeshie8
  • ganeshie8
are you allowed to use kmap ?
anonymous
  • anonymous
Professor hasn't taught it yet, only theorems http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf
ganeshie8
  • ganeshie8
Alright, its going to be tricky, lets see...
anonymous
  • 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
  • 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
  • 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
  • anonymous
How did you get that?
anonymous
  • anonymous
oh wait you are looking at the answer
anonymous
  • 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
  • anonymous
So the problem I'm simplifying is (W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)
ganeshie8
  • ganeshie8
Ohk.. I thought your started with WY'X + WY'Z' + W'X'Y + W'X'Z lets start over
ganeshie8
  • 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
  • anonymous
How did you applied X ( X + Y ) = X on step 2?
ganeshie8
  • ganeshie8
call it U(U+V) = U U = W+X' V = Z'
anonymous
  • anonymous
ooooo magic
ganeshie8
  • ganeshie8
not really we're just using the given theorems
anonymous
  • anonymous
(W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)
anonymous
  • anonymous
(W'+Y')(W'+X+Z')(W+X')(W+Y+Z) (brain still loading trying to figure out next step)
ganeshie8
  • 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
  • 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
  • anonymous
(W′+Y′)(W′+X+Z′) U+BC = (U+B)(U+C) U=W' B=Y' C=X+Z' W'+Y'(X+Z')
ganeshie8
  • ganeshie8
Yes
ganeshie8
  • ganeshie8
finally use theorem 16
anonymous
  • anonymous
W+X′(Y+Z) = (W+X')(Y+Z) right?
anonymous
  • anonymous
since W+X'(Y+Z) is way different and idk how to apply theorem 16 to it :[
ganeshie8
  • 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
  • anonymous
How did you applied it? (( X + Y ) ( X' + Z ) = X Z + X' Y)
ganeshie8
  • ganeshie8
(U+V)(U'+T) = UT + U'V U = W V = X'(Y+Z) T = Y'(X+Z')
anonymous
  • 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
  • ganeshie8
you need to understand well why those theorems work and convince everything in terms of conjunctions and disjunctions
ganeshie8
  • ganeshie8
for example, can you explain why below holds ? X + X' = 1
anonymous
  • anonymous
|dw:1441253021508:dw| my brain went completely blank when I was doing that
anonymous
  • anonymous
no, im unsure
ganeshie8
  • ganeshie8
you're confusing between AND gate and OR gate
ganeshie8
  • ganeshie8
X + X' is read as X or X' |dw:1441253525157:dw|
anonymous
  • anonymous
ow yes, my bad
ganeshie8
  • ganeshie8
it spits out 1 when at least one of its inputs is 1
ganeshie8
  • ganeshie8
easy to see that one of the X or X' is always 1
ganeshie8
  • ganeshie8
because X, X' are opposites of each other if X is 0, then X' will be 1 and viceversa
anonymous
  • anonymous
Why is it that X+0=X is not = 1?
ganeshie8
  • ganeshie8
X + 0 let X = 0 what does previous expression evaluate to
anonymous
  • anonymous
If X = 0 then we have 0 + 0 which is = 0
ganeshie8
  • ganeshie8
so X + 0 = 1 does not always hold

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