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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 ?

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    $\begingroup$ And the "even more right" notation is $p\to q$. $\endgroup$ May 14, 2013 at 9:35
  • $\begingroup$ I believe if that's the more right notation, this is so for intuitionist logic, only. $\endgroup$
    – Hibou57
    Jul 29, 2014 at 16:04

3 Answers 3

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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.

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    $\begingroup$ Thank you for sharing this with me, this was very interesting to read! $\endgroup$
    – Belgi
    May 14, 2013 at 15:19
  • $\begingroup$ This historical account is very illuminating! $\endgroup$
    – davyjones
    Feb 3 at 13:09
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One way to rationalise this notation is to think of a proposition as having a certain information content.

Then $p \supset q$ can be thought of as "the information content of $q$ is contained in that of $p$". A particular piece of information that can be obtained from $q$ is "$q$ is true".

Thus we see that $p \supset q$ naturally gives rise to the statement $p \implies q$, and conversely.


However, I agree with Hagen von Eitzen's comment that $p \to q$ should be the notation of choice, at least in symbolic logic. One could still use $\implies$ in a meta-context, i.e. as more or less a part of the natural language we discuss mathematics in.

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  • $\begingroup$ Thanks, now the notation is clear. Also thank you for explaining the difference between the two arrows notation $\endgroup$
    – Belgi
    May 14, 2013 at 10:10
  • $\begingroup$ Why is “$\rightarrow$” a better choice than “$\Rightarrow$” ? I see “$\Rightarrow$” everywhere… $\endgroup$
    – Hibou57
    Jul 29, 2014 at 16:13
  • $\begingroup$ I answered myself here: math.stackexchange.com/a/881914/115275 . $\endgroup$
    – Hibou57
    Jul 29, 2014 at 20:51
  • $\begingroup$ It is rather the other way around. From residuated lattices we know p≤q->r iff p^q≤r. Now take the special case p=1, we then have q->r iff q≤r. en.wikipedia.org/wiki/Residuated_lattice $\endgroup$
    – user4414
    Aug 23, 2014 at 14:42
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Notice that in the line in the link numbered 2, they say the symbol $p\supset q$ is often confused for set inclusion. But it is not set inclusion. It's just a symbol used to denote material implication. So, it seems you just exhibited how often it actually happens that this confusion arrises.

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  • $\begingroup$ Thanks I got confused. what I explained is something I saw when I took the basic set theory course $\endgroup$
    – Belgi
    May 14, 2013 at 10:11

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