Boolean Algebra help

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Boolean Algebra help

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

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Other answers:

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 )
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)
I mean I would wolfram...I don't think your teacher will mind. At this level we know how to distribute.
must show all work
are you allowed to use kmap ?
Professor hasn't taught it yet, only theorems http://ece224web.groups.et.byu.net/reference/boolean_algebra.pdf
Alright, its going to be tricky, lets see...
(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)
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)
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
\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) step1 : rearrange the product \((W+X'+Z')(W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\) step2 : use 10D on first two terms \((W+X')(W'+Y')(W'+X+Z')(W+Y+Z)\)
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)
\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) 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 \([W+X'(Y+Z)][W'+Y'(X+Z')]\)
(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 :[
\((W+X'+Z')(W'+Y')(W'+X+Z')(W+X')(W+Y+Z)\) 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 \(WY'(X+Z') + W'X'(Y+Z)\)
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')
What's your thought process when figuring out what theorem to use? I'm having a hard time figuring out what theorem to apply.
you need to understand well why those theorems work and convince everything in terms of conjunctions and disjunctions
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

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