De Morgan's Law Is it correct to say that de Morgan's Law is one of an isomorphism of classical logic?
I think it is.
(A bit meta, but is this question an appropriate one for this site?)
 A: You're basically right, but "an isomorphism of classical logic" is a bit vague. An isomorphism is between two objects. De Morgan's Laws provide an isomorphism between a Boolean algebra and its dual.
Let $B$ be a set and $\lor$ and $\land$ binary operations on $B$, and assume that for some unary operation $\neg$ on $B$ de Morgan's Laws hold, i.e.
$$
\begin{align}
\neg(a\lor b)=(\neg a)\land(\neg b)\;,
\\
\neg(a\land b)=(\neg a)\lor(\neg b)\;.
\end{align}
$$
for all $a,b\in B$. Then if $f:B\to B$ with $f(b)=\neg b$ is bijective, it provides an isomorphism between $(B,\lor,\land)$ and $(B,\land,\lor)$ in the sense that it is a bijection that respects the respective binary operations. In particular, if $(B,\lor,\land)$ is a Boolean algebra, then its dual $(B,\land,\lor)$ is also a Boolean algebra, and de Morgan's Laws provide an isomorphism between the two.
A: It definitely is an isomorphism. Of interest is that ¬ is its own inverse. Once you prove one of De Morgan's statements, you have established the isomorphism, and the other statement follows immediately, since it is just a statement of the isomorphism going in the other direction.  In fact, if you have an equivalence relation using only ¬, ∨, and ∧, you can negate both sides of the equivalence and, by repeated use of De Morgan's laws, end up with an equivalence identical to the original, but with ∨ and ∧ reversed.  Of course all the variable will be negated, but that is the same as having none of them negated.  As an example, once you have proved one of the distributive laws for ∨ and ∧, the other follows immediately.
