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anonymous
 one year ago
Boolean Algebra help
anonymous
 one year ago
Boolean Algebra help

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How 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
 one year ago
Best ResponseYou've already chosen the best response.0I tried distributing but it just make it worse and lengthy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0I mean X*X = X and X*X'=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's different than the other math you are thinking :[

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the dot which looks like multiplication is not multiplication. Stands for AND

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean OR! and the + stands for AND

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess I'll take the long path

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(WW′+WY′+X′W′+X′Y′+ZW′+ZY′)(W′+X+Z′)(W+X′)(W+Y+Z)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Expand it and then use that identity rule. Did that not cut many things down?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you want me to distribute it all? (working on it)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I mean I would wolfram...I don't think your teacher will mind. At this level we know how to distribute.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4are you allowed to use kmap ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Professor hasn't taught it yet, only theorems http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Alright, its going to be tricky, lets see...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.0W'(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
 one year ago
Best ResponseYou've already chosen the best response.4WY'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
 one year ago
Best ResponseYou've already chosen the best response.0How did you get that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait you are looking at the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0So the problem I'm simplifying is (W+X′+Z′)(W′+Y′)(W′+X+Z′)(W+X′)(W+Y+Z)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ohk.. I thought your started with WY'X + WY'Z' + W'X'Y + W'X'Z lets start over

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0How did you applied X ( X + Y ) = X on step 2?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4call it U(U+V) = U U = W+X' V = Z'

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4not really we're just using the given theorems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(W+X′)(W′+Y′)(W′+X+Z′)(W+Y+Z)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(W'+Y')(W'+X+Z')(W+X')(W+Y+Z) (brain still loading trying to figure out next step)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.0(W′+Y′)(W′+X+Z′) U+BC = (U+B)(U+C) U=W' B=Y' C=X+Z' W'+Y'(X+Z')

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4finally use theorem 16

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0W+X′(Y+Z) = (W+X')(Y+Z) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since W+X'(Y+Z) is way different and idk how to apply theorem 16 to it :[

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0How did you applied it? (( X + Y ) ( X' + Z ) = X Z + X' Y)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4(U+V)(U'+T) = UT + U'V U = W V = X'(Y+Z) T = Y'(X+Z')

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's your thought process when figuring out what theorem to use? I'm having a hard time figuring out what theorem to apply.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you need to understand well why those theorems work and convince everything in terms of conjunctions and disjunctions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4for example, can you explain why below holds ? X + X' = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441253021508:dw my brain went completely blank when I was doing that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you're confusing between AND gate and OR gate

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4X + X' is read as X or X' dw:1441253525157:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4it spits out 1 when at least one of its inputs is 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4easy to see that one of the X or X' is always 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4because X, X' are opposites of each other if X is 0, then X' will be 1 and viceversa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is it that X+0=X is not = 1?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4X + 0 let X = 0 what does previous expression evaluate to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If X = 0 then we have 0 + 0 which is = 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4so X + 0 = 1 does not always hold
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