# Duality in Disjunctive Normal Form and Conjunctive Normal Form

In my understanding, duality $$s ^{*}$$ of a compound proposition $$s$$ is obtained by replacing each $$\vee$$ to $$\wedge$$, each $$\wedge$$ to $$\vee$$, each $$T$$ to $$F$$, and each $$F$$ to $$T$$.

I have been looking into cases where $$s = s ^{*}$$, but I have found an interesting connection between duality and disjunctive normal form (DNF) and conjunctive normal form (CNF) while doing so.

For instance, suppose the DNF and the CNF of an unknown compound proposition is found through this truth table:

| p | q | ? |
|---|---|---|
| T | T | F |
| T | F | T |
| F | T | F |
| F | F | T |


Where,

DNF is $$(p \wedge \neg q) \vee (\neg p \wedge \neg q)$$

CNF is $$(\neg p \vee \neg q) \wedge (p \vee \neg q)$$

But this is where I found an interesting connection. The DNF and CNF are dual of each other.

In fact, whenever I find compound propositions where $$s = s ^{*}$$, its DNF and CNF are always dual of each other.

So, my question is, is this true (i.e. is there a counterexample)? If it is true, could $$s = s ^{*}$$ be also verified through finding the duality of $$s$$'s DNF and CNF, and conversely? Is there a name for this connection between duality and DNF and CNF?

Take the truth table and select all rows with output value $$0$$.
The input sections of these rows can be seen as DNF minterms of the inverse function.
If you then invert all literals, you get the CNF clauses of the original function.
None of the input literals of such a clause may be inverted without causing the output value to become $$0$$.
Apart from their ordering of minterms and clauses, neither DNF nor CNF of a function are necessarily unique. Therefore, they are no duals in a general sense.