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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?)

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up vote 2 down vote accepted

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.

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Thanks, this clears it up for me. – Casquibaldo Oct 17 '12 at 0:35

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