dc proof presents. the barber paradox a mathematical analysis does he, or doesnt he?

DC Proofwww.dcproof.com

Presents

The Barber Paradox

A Mathematical Analysis

Does he, or doesn’t he?

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

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!”

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!!

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 )

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

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

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

•Click the Premise button

•Enter “barber ε men”

barber ε men

•Click Continue

barber ε men

1 barber ε men Premise

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”

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…

1 barber ε men Premise

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

•Click the Specification button

1 barber ε men Premise

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

•Click line 2Specify: Click any active quantified statement

1 barber ε men Premise

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

•Enter “barber”

barber

1 barber ε men Premise

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

•Click Continue

barber

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

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

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

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 3Detach: Click any active IMPLICATION (=>) statement

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 1Detach: Click any active statement that mathces the LHS of line 3

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

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!

ConclusionOn 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 shavesDetach, 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…

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

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

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

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…”

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…”

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!

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

DC Proofwww.dcproof.com