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This question already has an answer here:

This is the informal proof of Drinker's paradox

The proof begins by recognising it is true that either everyone in the pub is drinking (in this particular round of drinks), or at least one person in the pub isn't drinking.

On the one hand, suppose everyone is drinking. For any particular person, it can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking — because everyone is drinking.

Suppose, on the other hand, at least one person isn't drinking. For that particular person, it still can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking — because that person is, in fact, not drinking.

I can agree with the first case of the proof. But how is the second case true ? How can they apply material implication in the second case when the material conditional itself has not been proved yet or given to us ?

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marked as duplicate by hardmath, JohnD, JimmyK4542, gnometorule, Lord_Farin Dec 24 '14 at 22:59

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    $\begingroup$ see math.stackexchange.com/questions/412387/… The part that confuses people is that if $P$ is a false conditional then $P\Rightarrow Q$ returns true. In this case, because the not drinker,$x$, isn't in fact drinking (the statement "$x$ is drinking" is false) any statement beginning with "if $x$ is drinking then" returns true. $\endgroup$ – JMoravitz Dec 22 '14 at 1:46
  • $\begingroup$ @JMoravitz Thanks, I think I get it now. So in the second case, they conclude the fact that $P \implies Q$ is true given the fact that not P is true ? $\endgroup$ – Sibi Dec 22 '14 at 1:51
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"If 1+1=3, then I am the King of France" is an example of a "vacuously true" statement - in classical logic, we take "P -> Q" as equivalent to "Q or !P". In this case "!P", so "Q or !P" is true (regardless of the truth value of Q), and so "P -> Q" is true; although this tells us little we didn't know before!

Similarly, in the second case you note, there exists a person A for whom "A is drinking" is false. Therefore we can take "A is drinking -> everyone is drinking" as equivalent to "(everyone is drinking) or !(A is drinking)" which is true, since "!(A is drinking)" is true regardless of whether (everyone is drinking) is true or not.

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  • $\begingroup$ I suppose you are French? $\endgroup$ – N.S.JOHN Oct 10 '16 at 11:08
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I would start at another way,

Start with an other but similar problem:

Is the statement "If the last person has left the room then everybody has left the room" true?

(who would disagree? )

So "there is some person and when that person has left the room then everybody has left the room"

and then replace "leaving the room" with "drinking"

and you are done

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