Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

Bertrand Russell Is the Pope (2010)

ceadserv1.nku.edu

Post reply on HN
Print view View original

31–40 of 122 posts

Re: Bertrand Russell Is the Pope (2010)

#31

Post by repsilat »

sinsterizme wrote: post #21
repsilat wrote: post #5

A more traditional argument might look something like Axiom 1: ((1=0 && 1!=0) -> Pope(Russell) Proposed axiom 2: (1=0 && 1!=0) 1&2 modus ponens: Pope(Russell)

Can you explain why Axiom 1 is an axiom? I don't understand this

I mostly just figured it was plausible-looking. The axiom-scheme could be "(X && !X) -> Y" for any expressions X and Y. Basically a partial definition of logical implication. The expression LHS->RHS is true when LHS is false, so it should at least be a theorem when the LHS is a contradition, and maybe an axiom depending on your taste. I suspect common taste is for small sets of axioms that don't tend to include this statement/scheme though.

Using the rules of inference from here[1], plus an assumed starting point for arithmetic, another deduction might be

1. Proposed axiom: (1==0)

2. Maybe some arithmetic axiom, humour me: !(1==0)

3. 2, negation elimination: (1==0) -> Pope(Russell)

4. 1, 3 modus ponens: Pope(Russell).

That "negation elimination" rule of inference looks kinda like what I did earlier I guess, but less questionable.

1: https://en.wikipedia.org/wiki/Propositional_calculus#Example...

Re: Bertrand Russell Is the Pope (2010)

#32

Post by DangitBobby »

I don't see how the conclusion must be that he's the Pope. It containing just him and the Pope but actually only just him is a contradiction that could just as easily be resolved by saying "it contains me and the Pope but actually just me so there is no Pope." Even if I accept the false proposition, the conclusion doesn't necessarily follow.

Re: Bertrand Russell Is the Pope (2010)

#33

Post by chmod600 »

perl4ever wrote: post #26
chmod600 wrote: post #18

Quotation (related only because it's from Russell): "The fundamental argument for freedom of opinion is the doubtfulness of all our belief... when the State intervenes to ensure the indoctrination of some doctrine, it does so because there is no conclusive evidence in favour of that doctrine .. It is clear that thought is not free if the profession of certain opinions make it impossible to make a living." https://en.…

conclusive evidence

Can you please expand on your point?

Re: Bertrand Russell Is the Pope (2010)

#34

Post by _ZeD_ »

DangitBobby wrote: post #32

I don't see how the conclusion must be that he's the Pope. It containing just him and the Pope but actually only just him is a contradiction that could just as easily be resolved by saying "it contains me and the Pope but actually just me so there is no Pope." Even if I accept the false proposition, the conclusion doesn't necessarily follow.

it's not that difficult to follow:

    let B = {set made of "Bertrand Russell" }, P = {set made of the pope}
    let BP = B ∪ P

    |B| = 1, |P| = 1 (trivials)
    |BP| may be 1 or 2
    if |BP| = 1 it means B=P -> Bertrand Russell is the Pope
    if |BP| = 2, and 2=1, it means B=P -> Bertrand Russell is the Pope

Re: Bertrand Russell Is the Pope (2010)

#35

Post by rembrandt »

_ZeD_ wrote: post #34
DangitBobby wrote: post #32

I don't see how the conclusion must be that he's the Pope. It containing just him and the Pope but actually only just him is a contradiction that could just as easily be resolved by saying "it contains me and the Pope but actually just me so there is no Pope." Even if I accept the false proposition, the conclusion doesn't necessarily follow.

it's not that difficult to follow: let B = {set made of "Bertrand Russell" }, P = {set made of the pope} let BP = B ∪ P |B| = 1, |P| = 1 (trivials) |BP| may be 1 or 2 if |BP| = 1 it means B=P -> Bertrand Russell is the Pope if |BP| = 2, and 2=1, it means B=P -> Bertrand Russell is the Pope

But since |BP| is both 1 and 2 it means Bertrand Russell is both the Pope, and not the Pope.

Re: Bertrand Russell Is the Pope (2010)

#36

Post by caspper69 »

bckr wrote: post #23
chmod600 wrote: post #18

Quotation (related only because it's from Russell): "The fundamental argument for freedom of opinion is the doubtfulness of all our belief... when the State intervenes to ensure the indoctrination of some doctrine, it does so because there is no conclusive evidence in favour of that doctrine .. It is clear that thought is not free if the profession of certain opinions make it impossible to make a living." https://en.…

And that will be used by shrill persons to claim censorship, etc, etc. However, I claim that the profession of certain opinions should make it impossible to make a living. EDIT: for certain values of "profess", which I'll expand as "go around telling everyone you frequently converse with"

What happens when it's your opinions that are unpopular?

Re: Bertrand Russell Is the Pope (2010)

#37

Post by sneak »

caspper69 wrote: post #36
bckr wrote: post #23

Earlier quoted context omitted.

And that will be used by shrill persons to claim censorship, etc, etc. However, I claim that the profession of certain opinions should make it impossible to make a living. EDIT: for certain values of "profess", which I'll expand as "go around telling everyone you frequently converse with"

What happens when it's your opinions that are unpopular?

Opinions are not innate, can shift, and are often wrong.

Re: Bertrand Russell Is the Pope (2010)

#38

Post by unanswered »

rembrandt wrote: post #35
_ZeD_ wrote: post #34

Earlier quoted context omitted.

it's not that difficult to follow: let B = {set made of "Bertrand Russell" }, P = {set made of the pope} let BP = B ∪ P |B| = 1, |P| = 1 (trivials) |BP| may be 1 or 2 if |BP| = 1 it means B=P -> Bertrand Russell is the Pope if |BP| = 2, and 2=1, it means B=P -> Bertrand Russell is the Pope

But since |BP| is both 1 and 2 it means Bertrand Russell is both the Pope, and not the Pope.

... yes. That is indeed what is meant by ex falso quodlibet. It does not make the example invalid.
Post reply on HN