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It seems to me that having ~(A AND B) = (~A) OR (~B) is obviously an isomorphism, but a Web search did not return anything. I am not sure how useful this is to know, but one thing that it does it to make each of the two DeMorgan laws a direct consequence of the other, making it necessary to directly prove only one of them. ################################################################
coffeemath 4: The point that I am trying to make is that it is the isomorphism that leads to duality. Consider the isomorphism between multiplication and addition: log(xy)=log(x)+log(y). Going in the reverse direction, we use exp(x), the inverse of log, to get exp(x+y)=exp(x)exp(y).

In the case of Boolean algebra, we get the isomorphism ~(X AND Y)= (~X) OR (~Y). NOT is self-inverse, so in the reverse direction we get ~(X OR Y) = ~ X AND ~Y. Because we use ~ in both directions, we can apply both rules to the same expression: We start S1 = (A OR (B AND C))
~S1 = ~(A OR (B AND C)) =
~A AND ~(B AND C) =
~A AND (~B OR ~C)), which is the dual of what we started with.

user1153980
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    You may want to look at this question: https://math.stackexchange.com/questions/408409/duality-principle-in-boolean-algebra – coffeemath May 17 '23 at 16:09
  • Thanks for the link. What needs to be said is that the duality principle follows from the DeMorgan Theorem isomorphism plus the fact that NOT is self-inverse, which makes it possible to negate any combination of ANDs, ORs and NOTs, so if you have some expression S1 = S2, then NOT S1 = NOT S2 and you can now use Demorgan's rules to get: dual of S1 = dual of S2. – user1153980 May 17 '23 at 17:06
  • Your later explanation included in your question looks OK to me. – coffeemath May 18 '23 at 05:59

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