TITLE: Using $p\supset q$ instead of $p\implies q$
QUESTION [8 upvotes]: I saw that a use for the notation $p\supset q$ instead of $p\implies q$
that got me a bit confused.
One occurrences is in this Wikipedia link.
It seems to me opposite than what it should be, let me explain what
I mean:
If $A,B$ are sets s.t $A\subset B$, $p$ is for $x\in A$, and $q$
is for $x\in B$ then we can identify (in some way) $$A\text{ with }p$$
$$B\text{ with }q$$
We have it that $p\implies q$, since $A\subset B$, but in the above
notation we have $q\subset p$ which to me looks like $B\subset A$
which is the opposite of what we wanted to express.
Can someone please explain to me the logic behind this notation ?

REPLY [12 votes]: Actually, historically, the reason we use $\supset$ is that Peano originally wrote $p C q$ for "$p$ is a consequence of $q$", and wrote a backward "$C$" for "$p$ has as a consequence $q$". Eventually, just as the "$\epsilon$" became "$\in$", so too did the backward "$C$" become "$\supset$".  So it doesn't actually have anything to do with set theory, per se, though maybe he used the same letter for both superset and consequence. 
But as Lord_Farin notes, one way to look at things might be in terms of "informational content" or even (very informally) "provability power" (if "$p \supset q$" is true, then accepting "$p$" requires accepting everything that would go with accepting "$q$"; though what I just said might be very philosophically contentious!). And yes, it seems more natural to think of it semantically, and so think that the "$\supset$" should be a "$\subset$". That's math for you.