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PublicouVictorhugo Gomes Alterado mais de 2 anos atrás

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DC Proof Presents

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The Barber Paradox A Mathematical Analysis Does he, or doesnt he?

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The Story The Mayor of the old town of Beardless was alarmed about a growing trenda beard growing trend!among the younger men in town In keeping with the glorious traditions their forefathers, the Mayor decreed that a man in town would be appointed the official Barber And that, for every man in Beardless, the Barber would be required by law to shave those and ONLY those men who do not shave themselves

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The Barber Sets Up Shop Every morning, the Barber would diligently shave those and ONLY those men who did NOT shave themselves But the Barber did not shave himself, for if he did, he would be breaking the law by shaving a man who DOES shave himself The law is the law, sir!

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Until, one day… The town constable burst into the Barbers shop and arrested him for dereliction of his official duties! The Barber, he said, had consistently failed to shave a man in Beardless who did NOT shave himself none other than the Barber himself!!

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What was to be done? How was this possible? How can this injustice be undone? To answer these questions, we will need to do a mathematical analysis of this bizarre situation using the DC Proof assistant (available free at )www.dcproof.com

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Why a mathematical analysis? The Barber Paradox arises out of the ambiguity of natural language What seems like a reasonable requirement can lead to an impossible situation, as here After translating it into the precise language of mathematics the resolution of this paradox will become immediately apparent

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Objects, Sets and Relations We will begin our analysis by identifying the following objects : –barber : the official town barber in Beardless –men : the set of all men in Beardless –shaves : a relation (a set of ordered pairs) defined on the set of men in Beardless Examples barber ε men means the barber is an element of the set of all men in Beardless; that is, the barber is a man in Beardless ( barber, x ) ε shaves means the barber shaves x ( x, x ) ε shaves means x shaves himself ~( x, x ) ε shaves means x does not shave himself

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We begin by opening DC Proof and introducing the initial premise that the barber is a man in the village…

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Click the Premise button

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Enter barber ε men barber ε men

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Click Continue barber ε men

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1 barber ε men Premise

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Using DC Proof… Applying the Premise Rule again, we suppose further that shaves is a relation such that for every man in Beardless, the barber shaves him if and only if he does not shave himself: 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise ALL(x) means for all x … => means implies means if and only if ~ means not

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Does the barber shave himself? On line 2, we have the definition of shaves : 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves] Premise Using the Universal Specification Rule, we can apply this definition to the barber himself…

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise Click the Specification button

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise Click line 2 Specify: Click any active quantified statement

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves] Premise Enter barber barber

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves] Premise Click Continue barber

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2

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On line 1, we have: 1 barber ε men Premise On line 3, we have: 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 Since the barber is a man in Beardless (line 1), we can apply the Detachment Rule to lines 1 and 3

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 Click the Detachment button

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 Click line 3 Detach: Click any active IMPLICATION ( => ) statement

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 Click line 1 Detach: Click any active statement that mathces the LHS of line 3

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1

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We have a contradiction! On line 4, we now have... (barber,barber) ε shaves ~(barber,barber) ε shaves If the barber shaves himself, then he must not shave himself If the barber does not shave himself, then he must shave himself Even if the barber is the only man in Beardless, this is an impossible requirement he must both shave himself and not shave himself! If we assume the existence of a relation shaves, as we have defined it here, then we will obtain the above contradiction Therefore, no such relation can exist!

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Conclusion On line 2, we have the Premise: 2ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise On line 4, we have the contradiction: 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 Since this statement is a contradiction, we will obtain the negation of the Premise on line 2 when we invoke the Conclusion Rule…

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 Click the Conclusion button

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 Optionally, change the bound variable shaves

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 Click Continue shaves

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 5 ~EXIST(shaves):ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Conclusion, 2 EXIST(x): means there exists an x such that…

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1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Premise 3 barber ε men => [(barber,barber) ε shaves ~(barber,barber) ε shaves] U Spec, 2 4 (barber,barber) ε shaves ~(barber,barber) ε shaves Detach, 3, 1 5 ~EXIST(shaves):ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] Conclusion, 2 ~EXIST(x): means there does not exist an x such that…

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In Conclusion… There does not exist a relation shaves such that, for every man in Beardless, the barber shaves those and only those men that do not shave themselves; that is… ~EXIST(shaves):ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] No combination of shavers and shaved can possibly satisfy the conditions set by this law! Therefore, the barber must go free!

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In Conclusion… There does not exist a relation shaves such that, for every man in Beardless, the barber shaves those and only those men that do not shave themselves; that is… ~EXIST(shaves):ALL(x):[x ε men => [(barber,x) ε shaves ~(x,x) ε shaves]] No combination of shavers and shaved can possibly satisfy the conditions set by this law! Therefore, the barber must go free! Artwork by Anna Vasilkova The End

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DC Proof

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