
#1
Nov1912, 05:22 AM

P: 87

Can somebody check if I answer them right?
Write the dual of the following boolean expressions: 1.x(y+ z)+z = x'+(yz')+z 2.x(y+ z)y = x+(y'z')+y' 3.xy+ yz+xz = (x+y)(y+z)'(x+z) 



#2
Nov1912, 05:41 AM

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P: 4,707





#3
Nov1912, 05:57 AM

P: 87





#4
Nov1912, 06:05 AM

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P: 4,707

duality of boolean expressionexpression such as (y+z)' Can you think of a way to confirm that you are applying that rule correctly? 



#5
Nov1912, 06:11 AM

P: 87





#6
Nov1912, 06:27 AM

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P: 4,707

Check expression #2.




#7
Nov1912, 06:41 AM

P: 87





#8
Nov1912, 07:02 AM

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P: 4,707

You really can't use an equals sign here: x(y+ z)+z = x'+(yz')+z because the two expression are NOT equal, nor are they meant to be equal. Perhaps type it as: x(y+ z)+z → x'+(yz')+z Or, even clearer: (x(y+ z)+z)^{D} = x'+(yz')+z so long as the reader is clear on what the superscript D denotes. 



#9
Nov1912, 07:07 AM

P: 87





#10
Nov1912, 07:42 AM

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P: 4,707

I think you'll find duality is of limited usefulness to you (except for answering exam questions). But where it can be applied is if someone goes to all the trouble of simplifying a complex Boolean expression to something equivalent, then, without any further mathematical effort, you can take their result and simply swap AND ↔ OR (and also swap constants 1 ↔ 0 ) and you'll arrive at another equation which you can with certainty say is also correct and justifying it by citing the principle of duality in Boolean algebra.. e.g., if I tell you that (a + b)' = a' . b' then without even understanding what it says you can write its DUAL and be confident that it also is a correct and valid Boolean equation, i.e., (a . b)' = a' + b' Well, that's my understanding anyway. Now, back to the problem at hand. I'd say unless you have been told to remove the complement outside the parentheses, you may as well leave it there, e.g., ( x.(y+ z) y )^{D} = x + (y . z)' + y' Have you been told you should remove the tick outside parentheses? Of course, if you want to remove it then apply De Morgan's theorem, as always. 


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