2. The Orwell Group

Those of you who have had an encounter with logic (logicians,mathematicians,philosophers,just interested) probably know Russell's paradox. If not, or if anyway, then the barber version goes as thus

*There does not exist a barber who shaves exactly those who does not shave themselves.*

To get a grip: say he exists, and then say he does not shave himself. But now he doesn't shave exactly those who does not shave them selves. Now suppose he does shave himself. But now he shaves someone who does shave himself. We can proof this using natural deduction:

all ([x] S b x => ~ S x x) premise; [p1] all ([x] ~ S x x => S b x) premise; [p2] S b b \/ ~ S b b by lem ; [lem1] ( S b b assumption; [h1] S b b => ~ S b b by all_e b p1 ; [res2] ~ S b b by imp_e h1 res2 ; [res3] bot by neg_e h1 res3 ) ; [dis1] ( ~ S b b assumption; [h1] ~ S b b => S b b by all_e b p2 ; [res2] S b b by imp_e h1 res2 ; [res3] bot by neg_e res3 h1 ) ; [dis2] bot by dis_e lem1 dis1 dis2 .Say that $b$ is our barber. We are quantifying over people, that is $x$. The predicate $S(x,y)$ is interpreted as "$x$ shaves $y$". So if $b$ encumber us with his existence, we can conclude absurdities. What a paradox.

I have heard this a lot: if the barber is female, the existence of such a barber is not absurd. Let me introduce $carol$, a female barber. Let $NST$ be the set of all them who *does not* shave themselves. Let $ST$ be the set of all them who *does* shave themselves. Let the statement be

$carol$ shaves *exactly* the set $NST$

By *exactly* we mean that $carol$ shaves everyone in $NST$, but none in $ST$. And we have

- $carol$ does
*not*shave herself. Hence $carol \in NST$ in which case $carol$ does not shave exactly the set $NST$. - $carol$ does shave herself. Hence we have $carol \in ST$ in which case $carol$ shaves someone in $ST$, and hence $carol$ does not shave exactly $NST$.

So the statement is still absurd. Gender do not matter to barber logic. Besides: do you really need a beard in order to be shaved?

We can go to whatever extremes. What if $carol$ has no legs or arms? She still does not shave exactly the set $NST$. What if $carol$ does not have a gender? Then *they* do not shave exactly $NST$. No matter $carol$ needs shaving at all, we still end up in one of the cases above. If $carol$ is hairless and feels shaving herself pointless and therefore doesn't do it, she still belongs to the set $NST$ and hence does not shave exactly $NST$. The only counterexample is the case where $carol$ besides misses her arms and legs has no body and no head. You know the case where $carol$ doesn't exists.

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