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Is the relation " being the contrapositive of" really symmetric?

I mean : the contrapositive of X --> Y is ~Y --> ~X.

If the relation " being the contrapositive of " is symmetric, then I can say that the contrapositive of : ~Y --> ~X is X --> Y.

But I ask myself whether, rigorously, the contrapostive of ~Y --> ~X should not rather be : 

                           ~ ~ X --> ~ ~ Y.

Certainly , ~ ~ X --> ~ ~ Y is equivalent to X --> Y. But is this equivalence a sufficient reason to say that the contraposition relation is symmetric?

Remark. The same question could be asked for the relation " being the nagation of". ~X is the negation of X , but is X the negation of ~X?

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    It depends on the details of the formulation. In the strictest sense you are correct, but if one considers not the formulas themselves, but the partition induced by considering sets of equivalent formulas, then the assertions are correct. – Git Gud May 14 '19 at 17:30
  • @MorganRodgers What are you talking about? The OP never questioned their equivalence. – Git Gud May 14 '19 at 20:16
  • Because, in fact, it the strictest sense (which happens to be a very reasonable context), the negation of $\neg X$ is $\neg \neg X$ which is different from $X$, despite them being equivalent. – Git Gud May 14 '19 at 21:48

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You are completely right, if we define the contrapositive as a syntactic transformation and don't want to do an awkward case analysis, then the contrapositive of $\neg X\to \neg Y$ is $\neg\neg Y\to \neg\neg X$.

This is equivalent to $Y\to X$ in classical logic but not, say, in intuitionistic logic where $\neg\neg X$ may differ from $X$.

Viewing $\neg X$ as $(X\to \bot)$ which is the usual definition of $\neg$ if it is not taken as a primitive, then the theorem "$(X\to Y)\to (\neg Y\to \neg X)$" requires basically no rules of logic. Of course, this is an exageration, but look :

Assume $X\to Y$, assume $\neg Y$, and assume $X$. Then by modus ponens, $Y$. Then by modus ponens [recall $\neg Y = (Y\to \bot)$] $\bot$. Thus $X\to \bot$ [discharge the $X$ assumption]. Therefore $\neg Y\to\neg X$ [discharge the $\neg Y$ assumption]. Therefore $(X\to Y)\to (\neg Y\to \neg X)$ [no assumptions].

And of course, this shows that $(\neg X\to \neg Y)\to (\neg\neg Y\to \neg\neg X)$ with also very few logic rules; whereas $(\neg X\to \neg Y)\to (Y\to X)$ requires a lot more : $\neg\neg Y\to \neg\neg X$ is the rightful contrapositive of $\neg X\to \neg Y$. (notice that if we defined $\neg_C X$ as $(X\to C)$ and the $C$-contrapositive of $X\to Y$ as $\neg_C Y\to \neg_C X$, then the theorem would still hold, whereas you wouldn't expect that $\neg_C X\to \neg_C Y$ implied $Y\to X$, would you ?)

Another point of view is to interpret (this is a very loose explanation) "$X\to Y$" as "functions from $X$ to $Y$". Then the contrapositive is nothing but functional application, whereas again, going from $\neg X\to \neg Y$ to $Y\to X$ requires more than that.

The answer for negation is the same : the negation of $\neg X$ is $\neg\neg X$, and it is equivalent to $X$ in classical logic, but even then syntactically it's different, and there are logics where the equivalence does not hold anymore.

Maxime Ramzi
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