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simplify the following using Boolean algebra. (xy)'+xz'+yz a)y+z' b)x+y c)X'+y d)xz'

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c) x'+y
$$ \Large{ (xy)'+xz'+yz\\ =[(xy)(xz')'(yz)']'\\ =[(xy)(x'+z)(y'+z')]'\\ =[((xyx')+(xyz))(y'+z')]'\\ =[0+xyzy'+xyzz']'\\ =[0+0+0]'\\ =[0]'\\ =1 } $$ This is a tautology, because it is always true. Here is another method using a Truth Table |dw:1395590669537:dw| This also shows a tautology
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Other answers:

(xy)' + xz' + yz We know that (xy)' = x' + y' [De morgan's laws] x' + y' + xz' + yz x' + z' + y' + z x' + y' + 1 = 1 None of the options are as 1 @ybarrap Is truth table the only method to find that out?
@moli1993 I guess you have to see which of the options also show Tautology [Always true].
The truth table is authoritative. This is a tautology. None of the options are a tautology so they are all incorrect.
I am so sorry , wrote the wrong question here :- it is (xz)'+xz'+yz and the options are same as given earlier !
It's the same process
process is same but it's not a tautology i suppose !
(xz)' = x' + z' z' +xz' = z'(1+x) = z' --> x' + z' + yz = x' +z' +z = x' + 1 = 1
how x' + z' + yz = x' +z' +z
(xz)'+xz'+yz We know that (xz)' = x' + z' [De morgan's laws] x' + z' + xz' + yz x' + (z' + xz') + yz [ In z' + xz' take z' common, z'(1+x) = z'(1) = z' ] x' + z' + yz [yz = zy] z' + zy + x' Using simplification law [A + A'B = A+B] we get, z' + y + x' Final answer should be x' + y + z'
but it doesnt matches any of the options given above
(xz)' + y I am pretty sure, this is the final answer. :/ @ybarrap @dumbcow ?
me too I think u r correct !
yep http://www.wolframalpha.com/input/?i=~%28x+and+z%29+or+%28x+and+~z%29+or+%28y+and+z%29
yes it is
Thank you all for helping me !
:)

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