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The barbers shave all those who do not shave themselves. Therefore, the barber shaves himself.

I need to write this in terms of quantifiers of mathematical logic. (b,Sxy)

"So far i have done:

EDIT

b-->(∀y(Sxy))

I am not sure how to write the conclusion

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2 Answers 2

I think probably you want to change your $Sxy$ slightly. One way that you can tell there is something wrong with it is that there is no mention of $y$ in its description!

Try re-writing it as: $Sxy$ means "$x$ shaves $y$"

Can you write everything out now?

[For all barbers: given any person who doesn't shave his own head, the barber shaves that person's head; therefore, the barber shaves his own head.]

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I think that the idea is to have a predicate $S(x,y)$ to mean $x\text{ shaves } y$, so to say that $x$ shaves himself is to say $S(x,x)$. Of course we will also have a predicate $B(x)$ which says that $x$ is a barber.

Now you want to have a barber that shaves all those who do not shave themselves. Namely, someone which is a barber, and for every person who does not shave themselves, the barber shaves them. You also want that all the barbers for which this property holds, would shave themselves.

So we have: $$\forall x\Big(\big(B(x)\land\forall y(\lnot S(y,y)\rightarrow S(x,y))\big)\rightarrow S(x,x)\Big)$$

So we said that for any $x$, if $x$ is a barber, and whenever $y$ does not shave himself, then $x$ shaves $y$, then it follows that $x$ shaves himself.

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Does the proposition imply that there exists such a barber? –  alancalvitti Nov 2 '12 at 15:02
    
Re my previous comment referred to the opening existential quantifier in the answer above - and which was subsequently edited to the universal. –  alancalvitti Nov 2 '12 at 16:31
    
@alan: Yes. Thank you. Just so you know, it is equally possible to just delete the previous comment instead of posting a comment explaining it after an edit. –  Asaf Karagila Nov 2 '12 at 16:49

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