Carregar apresentação

A apresentação está carregando. Por favor, espere

PublicouVictorhugo Gomes Alterado mais de 5 anos atrás

1
**DC Proof www.dcproof.com Presents**

Presenter’s Notes Some Background on the Barber Paradox Much has been written about the Barber Paradox, much of it not very illuminating. Many writers on the subject simply throw up their hands noting only that they have arrived at a contradiction. A necessary first step, but for every logical contradiction, there is an underlying false premise. Others have proposed changing the rules of the game, e.g. allowing the barber to be someone or something that is not a man in the town, e.g. a woman, an outsider or even a robot. Not a very satisfying “resolution” to say the least. And completely unnecessary, as we see here. (In the original narrative, the barber is a man in town who shaves those and only those men in town who do not shave themselves.) Some have proposed banning all forms of self-reference – shaving yourself is OUT! Some of the more logically minded have insisted that the barber or his customers cannot exist. And they can “prove” it. Not a bad approach, but it misses the point. Here, we see that the resolution of the Paradox lies not in changing the rules, banning self reference or the non-existence of the barber or his customers, but in the fact that the “system requirements” given in the narrative make it is impossible to consistently define who shaves whom. We see here that one of the implicit “system requirements” is that the barber both shaves himself and does NOT shave himself! No such relation can exist, of course. Uncritically assuming otherwise is the source of this seeming “paradox.” Using first-order-logic alone, we cannot prove the existence or non-existence of any relation. Using set theory (mathematics), however, we can prove the existence or, in this case, the non-existence of a relation (defined as a set of ordered pairs). Hence the need for a mathematical analysis. Outline Essentially, we prove: ALL(b):ALL(m):[b εm => ~EXIST(s):ALL(x):[x εm => [(b,x) εs <=> ~(x,x)εs]]] We begin by supposing that b ε m, then supposing further that: EXIST(s):ALL(x):[x εm => [(b,x) εs<=>~(x,x)εs]]. We then obtain the contradiction (b,b)εs <=> ~(b,b) εs. Therefore, ~EXIST(s):ALL(x):[x εm => [(b,x) εs<=>~(x,x)εs]]. Dan Christensen November 2011

2
**A Mathematical Analysis**

The Barber Paradox A Mathematical Analysis Does he, or doesn’t he?

3
The Story The Mayor of the old town of Beardless was alarmed about a growing trend─a 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

4
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!”

5
Until, one day… The town constable burst into the Barber’s 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!!

6
**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 )

7
**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

8
**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

9
We begin by opening DC Proof and introducing the initial premise that the barber is a man in the village…

11
**Click the Premise button**

12
barber ε men Enter “barber ε men”

13
barber ε men Click Continue

14
1 barber ε men Premise

15
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”

16
**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…

17
**Click the Specification button**

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

18
**Click line 2 1 barber ε men Premise ALL(x):[x ε men**

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

19
**Enter “barber” 1 barber ε men Premise 2 ALL(x):[x ε men**

=> [(barber,x) ε shaves <=> ~(x,x) ε shaves] barber Enter “barber”

20
**Click Continue 1 barber ε men Premise 2 ALL(x):[x ε men**

=> [(barber,x) ε shaves <=> ~(x,x) ε shaves] barber Click Continue

21
1 barber ε men Premise ALL(x):[x ε men => [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 3 barber ε men => [(barber,barber) ε shaves <=> ~(barber,barber) ε shaves] U Spec, 2

22
**=> [(barber,barber) ε shaves <=> ~(barber,barber) ε shaves]**

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

23
**Click the Detachment button**

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

24
**Click line 3 1 barber ε men Premise ALL(x):[x ε men**

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

25
**Click line 1 1 barber ε men Premise ALL(x):[x ε men**

=> [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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

26
1 barber ε men Premise ALL(x):[x ε men => [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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

27
**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!

28
**Conclusion On line 2, we have the Premise: 2 ALL(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…

29
**Click the Conclusion button**

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

30
**Optionally, change the bound variable**

1 barber ε men Premise ALL(x):[x ε men => [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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 shaves Optionally, change the bound variable

31
**Click Continue 1 barber ε men Premise ALL(x):[x ε men**

=> [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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 shaves Click Continue

32
**EXIST(x): means “there exists an x such that…”**

1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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 Conclusion, 2 EXIST(x): means “there exists an x such that…”

33
**~EXIST(x): means “there does not exist an x such that…”**

1 barber ε men Premise 2 ALL(x):[x ε men => [(barber,x) ε shaves <=> ~(x,x) ε shaves]] 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 Conclusion, 2 ~EXIST(x): means “there does not exist an x such that…”

34
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!

35
**Artwork by Anna Vasilkova**

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! The End Artwork by Anna Vasilkova

36
**DC Proof www.dcproof.com**

Presenter’s Notes Some Background on the Barber Paradox Much has been written about the Barber Paradox, much of it not very illuminating. Many writers on the subject simply throw up their hands noting only that they have arrived at a contradiction. A necessary first step, but for every logical contradiction, there is an underlying false premise. Others have proposed changing the rules of the game, e.g. allowing the barber to be someone or something that is not a man in the town, e.g. a woman, an outsider or even a robot. Not a very satisfying “resolution” to say the least. And completely unnecessary, as we see here. (In the original narrative, the barber is a man in town who shaves those and only those men in town who do not shave themselves.) Some have proposed banning all forms of self-reference – shaving yourself is OUT! Some of the more logically minded have insisted that the barber or his customers cannot exist. And they can “prove” it. Not a bad approach, but it misses the point. Here, we see that the resolution of the Paradox lies not in changing the rules, banning self reference or the non-existence of the barber or his customers, but in the fact that the “system requirements” given in the narrative make it is impossible to consistently define who shaves whom. We see here that one of the implicit “system requirements” is that the barber both shaves himself and does NOT shave himself! No such relation can exist, of course. Uncritically assuming otherwise is the source of this seeming “paradox.” Using first-order-logic alone, we cannot prove the existence or non-existence of any relation. Using set theory (mathematics), however, we can prove the existence or, in this case, the non-existence of a relation (defined as a set of ordered pairs). Hence the need for a mathematical analysis. Outline Essentially, we prove: ALL(b):ALL(m):[b εm => ~EXIST(s):ALL(x):[x εm => [(b,x) εs <=> ~(x,x)εs]]] We begin by supposing that b ε m, then supposing further that: EXIST(s):ALL(x):[x εm => [(b,x) εs<=>~(x,x)εs]]. We then obtain the contradiction (b,b)εs <=> ~(b,b) εs. Therefore, ~EXIST(s):ALL(x):[x εm => [(b,x) εs<=>~(x,x)εs]]. Dan Christensen November 2011

Apresentações semelhantes

OK

Aula Teórica 18 & 19 Adimensionalização. Nº de Reynolds e Nº de Froude. Teorema dos PI’s , Diagrama de Moody, Equação de Bernoulli Generalizada e Coeficientes.

Aula Teórica 18 & 19 Adimensionalização. Nº de Reynolds e Nº de Froude. Teorema dos PI’s , Diagrama de Moody, Equação de Bernoulli Generalizada e Coeficientes.

© 2019 SlidePlayer.com.br Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Anúncios Google