# Boolean algebra duality principle

Discussion in 'Homework Help' started by xEnOnn, Feb 3, 2011.

1. ### xEnOnn Thread Starter New Member

Feb 2, 2011
25
0
My textbook says that the duality principle of the boolean algebra allows me to obtain a same expression in a different form. So for example, X.Y+Z' = (X'+Y').Z

But when I try to plot the truth table, the values of X.Y+Z' is opposite of (X'+Y').Z, which is if one of it is true, the other is false and vice versa. Since this is the case, they are not the same right? Then why are they still considered equal?

2. ### Heavydoody Active Member

Jul 31, 2009
140
11
I am just now doing this at school, so I am no expert. Having said that, my understanding of duality is that for every axiom there is a "dual" axiom created by exchanging and's for or's, or's for and's, 1's for 0's, and 0's for 1's. These dual axioms do not equal each other. So, for example:

x+1=1 has the dual x0=0
(xy)'=x'+y' has the dual (x+y)'=x'y'

However, x+1≠x0 and (xy)'≠x'y'

3. ### narasimhan Member

Dec 3, 2009
72
6
Absolutely correct.

Actually the dual of X.Y+Z' is (X+Y),Z'
Moreover (X'+Y').Z is the complement of X.Y+Z'

that is ((X'+Y').Z)'=X.Y+Z'
Now draw truth table and verify. And do reply if you get it right and understood the concept.

4. ### xEnOnn Thread Starter New Member

Feb 2, 2011
25
0
hmm.. so the dual axioms are not equal of each other. but I am still a little confused.

So I don't use duality when simplifying expressions since they are not equivalent of each other?

And when given a SOP and asked to convert to a POS, it is just the complement of SOP right? So POS = (SOP)' right?

Funny thing is when I have say X'.Z' + W'.Z as a SOP, and when I "NOT" it to get the POS, I have this: (X'.Z' + W'.Z)' = (X+Z).(W+Z'). But the answer is (X'+Z).(W'+Z')

From X'.Z' + W'.Z to (X'+Z).(W'+Z') looks like duality but isn't the POS and SOP complements of each other rather than dual?thanks.

5. ### narasimhan Member

Dec 3, 2009
72
6
That's wrong
Well you got it all messed up. Please read Digital design by Morris Mano. It'll be hard for me to explain the concepts.
First get to know the difference between minterm and maxterm and then POS and SOP.

But anyway here you go
Minterm'=Maxterm

A function F is given in SOP form. to find the POS of F do the following steps.
1. Find the minterms of F
2. Find the SOP of F'(by combining the omitted minterms)
3. Take complement of F' i.e. (F')'=F So you would end up with the POS of F

And remember this
POS of F=(SOP of F')'

6. ### xEnOnn Thread Starter New Member

Feb 2, 2011
25
0
oohhh! The POS is NOT the complement of SOP.

I tried your steps and it works! I got the POS of function F from a simplified SOP of function F.

So the POS is the complement of the omitted minterms, which are the maxterms that return 0 on the truth table. Therefore, I cannot just complement the minterms that I have because by doing so, these are the maxterms that return 1 but not 0, is this right?

Just to clarify something, the minterm are just any m0, m1, m# terms that are not neccessary to need to have result of 1, right? It can also can a result of 0. It is just during the sum of minterms that we gather only the minterms that return 1, right? Similarly for maxterms, the results of the maxterm can either be 0 or 1. But the maxterms must have a result of 0 when I want to put it into a POS.

7. ### narasimhan Member

Dec 3, 2009
72
6
Till this I understand. You are making quite a progress.

I guess this is also right.

This is a bit confusing.
For sum of minterms you need to include all minterms which are 1.

For product of maxterms you need to include all maxterms which are 0.

Moreover you can't combine minterms and maxterms

8. ### xEnOnn Thread Starter New Member

Feb 2, 2011
25
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Say if a minterm of A.B'.C' that returns 0. It is not included in the canonical expression of a function F. Does it still being considered as a minterm of the function F even though it does not result in an 1 and is not included in the canonical expression?

So when I complement a minterm that returns 1, it will just be its maxterm that returns 1.

Is listing all the minterms of a simplified expression and the find its omitting minterms and then sum these omitting minterms and then complement them the only way to find the POS? I find myself taking quite long to complete this process.

I see some similarity in SOP to POS with duality. I am not sure if it is just out of coincidence but can it be used as a short cut? Since duality doesn't mean its "dual'ed" form isn't equals to its original form, how is it used in boolean simplification?

9. ### narasimhan Member

Dec 3, 2009
72
6
It is not a minterm of F

If you complement a minterm of F it'll become a maxterm of F'(not F).

It is the procedure. Once you master it you can do most of the steps in your mind without pencil and directly arrive at the POS. But it'll require lot of practice.

I'll give an example
You're need to prove the two properties
A) xy+x'z+yz=xy+x'z
B) (x+y)(x'+z)(y+z)=(x+y)(x'+z)

After you prove A you need not prove B. You can just say by duality principle B is also true. i.e B is the dual of A

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10. ### xEnOnn Thread Starter New Member

Feb 2, 2011
25
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ahh... okay... I get what the duality is for now. It is not for direct expression equivalence. but rather a way to use a rule that was proven for a SOP on a POS or vice versa.

thanks!! thank you so much!!

11. ### narasimhan Member

Dec 3, 2009
72
6
Ya. There are other ways to use it(but doesn't strike my mind).

12. ### zebloc New Member

Jul 17, 2013
1
1
"The Dual of an identity is also an identity. This is called the Duality Principle". A Boolean Identity is X+0=X or X+X=X. There's lots of them. Duals only work with identities. To find the Dual you switch operators (+ & .) and switch identity elements (0 & 1, if there are any 0's and 1's) to change X+0=X to X.1=X and to change X+X=X to X.X=X which creates new identities which are also valid. There is no meaning to creating a Dual from an arbitrary expression like X'Y+XY'=1. A Complement depends on an arbitrary expression like f1(x,y)=X'Y+XY', the complement of which would be f2(x,y)=(X+Y').(X'+Y) which if you plug values into f1(x,y) will give you the exact opposite results if the same values are plugged into f2(x,y). A Complement is formed by negating each variable and switching each operator.

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13. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
...you are talking to the ghosts of the forum... People typically try to stay away from matters of the occult...