Duality in propositional logic between conjunction and disjunction, $K$ and $A$ means that for any "identity", such as $KpNp = 0$ (ignoring the detail of how to define this notion in propositional logic), if we replace all instances of $K$ by $A$, all instances of $A$ by $K$, all instances of 1 by 0, and all instances of 0 by 1, the resulting equation will also consists of an "identity", $ApNp = 1$. Suppose that instead of conjunction "$K$" and disjunction "$A$", we consider any pair of "dual" operations $\{Y, Z\}$ of the 16 logical operations such that they qualify as isomorphic via the negation operation $N$, where $Y$ does not equal $Z$. By "isomorphic" I mean that the sub-systems $(Y, \{0, 1\})$, $(Z, \{0, 1\})$ are isomorphic in the usual way via the negation operation $N$, for example $K$ and $A$ qualify as "isomorphic" in the sense I've used it here.
If we have an identity involving operations $A_1, \dots, A_x$, and replace each instance of each operation by its "dual" $A'_1$, ..., $A'_x$, replace each instance of 1 by 0, and each instance of 0 by 1, is the resulting equation also an identity? If so, how does one prove this? How does one show that the equation obtained via the "duality" transformation here is also an identity?