duality of boolean expression

aruwin

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+ y’z’+xz = (x+y)(y+z)'(x+z)

NascentOxygen

........

aruwin

Oh, it should be x’(y+ z’)+z = x'+(yz')z. Now this is correct,right?

NascentOxygen

I surmise that you have been told of a rule you can apply when taking the dual of a complemented
expression such as (y+z)'

Can you think of a way to confirm that you are applying that rule correctly?

aruwin

Yes, by drawing the truth table.

NascentOxygen

Check expression #2.

aruwin

Ok, the results are not the same :( What should I do?

NascentOxygen

Actually, your expressions are a bit sloppy and I think you should be encouraged to observe more mathematical rigor.

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.

aruwin

Wait a minute. So, what you're saying is that dual expressions are not necessarily equal to each other???I thought they were always equal, it's just that we change or to and and vice versa, but the outputs are always the same. So, they're not actually equal?

NascentOxygen

They are not necessarily equal. In fact, I think you cannot make a general comparison.

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.