The Drinker’s Paradox: A Tale of Three Paradoxes (reposted)

[Dan Christensen]

The Drinker’s Theorem: Consider the set of all drinkers in the world, and the set of all people in a given pub. Then there exists a person who, if he or she is drinking, then everyone in that pub is drinking.

It doesn’t matter how many people are there. Or how many are drinking. Or how few. No one needs to be taking their cues from some “lead drinker,” but in every pub, in every town and village, it just happens! How is this possible?

There are several possible approaches to this problem. Here, we will turn to British philosopher and mathematician, Bertrand Russell (1872 – 1970). His famous Paradox is the key.

See: https://dcproof.wordpress.com/2014/06/03/the-drinkers-paradox/

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Johnnie Bertolini]

Dan Christensen wrote:

> The Drinker’s Theorem: Consider the set of all drinkers in the world,
> and the set of all people in a given pub. Then there exists a person
> who, if he or she is drinking, then everyone in that pub is drinking.

AFU Unit Describes Command Ordering Them To Shoot Every Civilian In Town
And Their Own Wounded
https://%62%69%74%63%68%75%74%65%2E%63%6F%6D/video/ZuIn2nrQ1DYt/

[Mostowski Collapse]

Is this some sort of reference that our wonky man is always drunk?
Especially when he claims he posses some "right" definition?

[Dan Christensen]

On Wednesday, October 5, 2022 at 5:24:12 PM UTC-4, Johnnie Bertolini wrote:
> Dan Christensen wrote:
>
> > The Drinker’s Theorem: Consider the set of all drinkers in the world,
> > and the set of all people in a given pub. Then there exists a person
> > who, if he or she is drinking, then everyone in that pub is drinking.

> AFU Unit Describes Command Ordering Them To Shoot Every Civilian In Town
> And Their Own Wounded

And you support this, Nazi boy???

So, have you been, ahem... "volunteered" to die for Putin's massive ego yet? It may be too late to flee Nazi ruSSia. Maybe go into hiding for a few years in a remote little dacha in the forest in the far north. It might be a nice change for you. Be careful who you talk to though. You never know who might turn you in.

[Dan Christensen]

On Wednesday, October 5, 2022 at 6:11:47 PM UTC-4, Mostowski Collapse wrote:
> Is this some sort of reference that our wonky man is always drunk?
> Especially when he claims he posses some "right" definition?

See my reply just now to your identical posting at sci.logic

Dan

[Cesario De rege]

Dan Christensen wrote:

> On Wednesday, October 5, 2022 at 5:24:12 PM UTC-4, Johnnie Bertolini
> wrote:
>> Dan Christensen wrote:
>>
>> > The Drinker’s Theorem: Consider the set of all drinkers in the world,
>> > and the set of all people in a given pub. Then there exists a person
>> > who, if he or she is drinking, then everyone in that pub is drinking.
>
>> AFU Unit Describes Command Ordering Them To Shoot Every Civilian In
>> Town And Their Own Wounded
>
> And you support this, Nazi boy???

this bad teeth idiot doesn't know what *AFU* is, pushing me fart into his
coffee, so he undrestand.

AFU Unit Describes Command Ordering Them To Shoot Every Civilian In Town
And Their Own Wounded

https://%62%69%74%63%68%75%74%65%2E%63%6F%6D/video/ZuIn2nrQ1DYt/

[Mostowski Collapse]

I really enjoy your DC Spoiled proof tool. It is no applications
in mathematics whatever so far. But its a funny contraption to

refute all your nonsense claims, wonky man!

LMAO!

[Mostowski Collapse]

This crankness of wonky man, is on par with AP brain farto,
who thinks his ellipses are oval. Now we have wonky man,
who thinks he doesn't need set theory, and he also thinks

set theory is nowhere used in mathematics. But on the
first problem of formalizing his saying:

Dan Christensen schrieb am Samstag, 1. Oktober 2022 um 20:33:11 UTC+2:
> If we are given only a function f mapping elements of set A to elements
> of set B, nothing can be inferred about f(x) from x not in A. In this case,
> we say that f(x) is UNDEFINED.
https://groups.google.com/g/sci.logic/c/InGcDaFcuKE/m/ZesxKIrSAAAJ

The result is a cloud of chaos with insane claims, like
Undefined(_,_,_,_) bla bla over Undefined(_,_), and we
see befor our eyes, cards houses erect and crashing

into itself, none of his silly depositions are true!

What is your scientific method, wonky man?

[Fritz Feldhase]

On Thursday, October 6, 2022 at 3:00:47 AM UTC+2, Mostowski Collapse wrote:

> This crankness of wonky man, is on par with AP brain farto,
> who thinks his ellipses are oval.

Well, actually ellipses ARE ovals. :-)

"An ellipse is always an oval, but an oval is not always an ellipse. (Ellipses are a subset of ovals)"
Source: https://www.differencebetween.com/difference-between-ellipse-and-vs-oval/

But AP claims that "his ovals" (i. e. certain cone sections) _aren't ellipses_.

[Dan Christensen]

On Wednesday, October 5, 2022 at 8:45:51 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> I really enjoy your DC Proof proof tool.

Thanks. You are getting pretty good at proving stuff, Jan Burse. Just not the right stuff. ;^) Don't be discouraged.

> It is no applications
> in mathematics whatever so far.

You are confusing math and what might be called "philosophy." DC Proof may not be of much use in your philosophy class. But your philosophy class isn't of much use in math. DC Proof is based on the methods of proof that are actually used by mathematicians in proofs---not usually the FOL or ZFC methods of your philosophy class as it turns out. Sorry.

[Chris M. Thomasson]

On 10/5/2022 1:12 PM, Dan Christensen wrote:
> The Drinker’s Theorem: Consider the set of all drinkers in the world, and the set of all people in a given pub. Then there exists a person who, if he or she is drinking, then everyone in that pub is drinking.
>
> It doesn’t matter how many people are there. Or how many are drinking. Or how few. No one needs to be taking their cues from some “lead drinker,” but in every pub, in every town and village, it just happens! How is this possible?

[...]

For some reason, in real life, there was always a person that would not
be drinking at a bar. Designated driver, or even the bartender.

[Fritz Feldhase]

On Thursday, October 6, 2022 at 3:53:04 AM UTC+2, Dan Christensen wrote:

> ... methods of your philosophy class ...

Seems that you know a lot more about "philosophy classes" that "about math classes". And it shows!

Actually, you are a complete nutjob concerning mathematics.

[Fritz Feldhase]

On Thursday, October 6, 2022 at 3:57:16 AM UTC+2, Chris M. Thomasson wrote:
> On 10/5/2022 1:12 PM, Dan Christensen wrote:
> >
> > The Drinker’s Theorem: Consider the set of all drinkers in the world, and the set of all people in a given pub. Then there exists a person who, if he or she is drinking, then everyone in that pub is drinking.

Consider an empty (closed) bar. Then everyone in that pub is drinking. That's for sure!

Actually, DC even mentions this possibility HERE:

"The Drinker's Theorem: Consider the set of all drinkers in the world, and

the set of all people in a given pub (empty or otherwise). Then [...]"

Source: https://www.dcproof.com/DrinkersParadox.html

But actually, the "theorem" is false. What if there aren't any persons? Then clearly there is no person "such that", since there simply IS NO person.

Actually, his explanation concerning his "The Drinker’s Theorem" is nonsense.

To be generous I even allow for the two unstated premises:

1. All drinkers [in the world] are persons.
2. All people [in a (any) pub] are persons.

Of course, if there are no persons, then the set of all drinkers in the word and the set of all people in a given bar are empty. So what?

We see that his "theorem" does not hold "in general". Sorry, Dan.

> For some reason, in real life, there was always a person that would not
> be drinking at a bar. Designated driver, or even the bartender.

Not if the bar is empty (closed), which may happen in real life. :-P

( Btw. _every_ element in the empty set is drinking too! )

[Archimedes Plutonium]

Fritz head up arse Feldhase, a square is a rectangle, square being a subset of rectangle. And Fritz would say-- but a rectangle is not always a square.

In the Navy they often said-- before opening the mouth, Fritz, screw on your brains first.

******* Real truth

Oval is a different curve altogether from a ellipse. Ellipse is a different curve altogether from a oval. These jackarse idiots like Feldhase and Baez before him, who want to sweep away their errors with subset. Are crazy loons that never belong in science.

They come on as saying apple is a subset of orange, or banana is a subset of zucchini.

[Chris M. Thomasson]

;^D

You are most likely correct. The last person closing the bar, has a big
drink of 80 proof and calls a cab. Well, then the bar would not
technically be empty until the last person actually leaves? There might
be a person or two who seem to sleep at the damn bar, they always seem
to be there... ;^)

> ( Btw. _every_ element in the empty set is drinking too! )

Well, I suppose if an element of an "empty set" (btw, that does not make
too much sense to me) is drinking... Well, please do not let the empty
set get into a damn car! Even if a non intoxicated element of the empty
set is driving. Also, somebody, another element of the empty set might
puke! ;^)

[Dan Christensen]

It's a cautionary tale. If you have an existential quantifier on an implication, you should expect some seriously weird results.

Compare (note bracketing):

1. EXIST(x):[x in drinkers => ALL(a):[a in pub => a in drinkers]] <---- Existential quantifier over an implication

2.EXIST(x):[x in drinkers] => ALL(a):[a in pub => a in drinkers] <---- Not an existential quantifier over an implication

(2) is clearly not always always true in real life.

Strangely, (1) is always true in classical logic and set theory.

Interestingly, for any set s and proposition Q, we have:

EXIST(x):[x in s => Q]

See link to proof at http://www.dcproof.com/DrinkersParadox.html

[Chris M. Thomasson]

On 10/5/2022 7:25 PM, Archimedes Plutonium wrote:
> On Wednesday, October 5, 2022 at 8:28:43 PM UTC-5, Fritz Feldhase wrote:
>> On Thursday, October 6, 2022 at 3:00:47 AM UTC+2, Mostowski Collapse wrote:
>>
>>> This crankness of wonky man, is on par with AP brain farto,
>>> who thinks his ellipses are oval.
>> Well, actually ellipses ARE ovals. :-)
>>
>> "An ellipse is always an oval, but an oval is not always an ellipse. (Ellipses are a subset of ovals)"
>> Source: https://www.differencebetween.com/difference-between-ellipse-and-vs-oval/
>>
>> But AP claims that "his ovals" (i. e. certain cone sections) _aren't ellipses_.
>
>
> Fritz head up arse Feldhase, a square is a rectangle, square being a subset of rectangle. And Fritz would say-- but a rectangle is not always a square.

[...]

Try to tone down the anger a bit? A circle is a special case of an
ellipse, right? For some reason, I think of an ellipse as a unit circle
rotated in 3d.

[Fritz Feldhase]

On Thursday, October 6, 2022 at 5:05:18 AM UTC+2, Dan Christensen wrote:

> 1. EXIST(x):[x in drinkers => ALL(a):[a in pub => a in drinkers]]
>

> Strangely, (1) is always true in classical logic and set theory.

Not that strange. Actually, the SET of drinkers certainly will not be a drinker (won't you think so)?

In other word's (in agreement with, say, ZFC set theory), ~(drinkers in drinkers).

In this case (vacuously): drinkers in drinkers => <whatever>

And hence especially: drinkers in drinkers => ALL(a):[a in pub => a in drinkers]

From this we get (in the context of classical logic, where all terms denote):

EXIST(x):[x in drinkers => ALL(a):[a in pub => a in drinkers]].

Not really a "paradox". And not really well expressed in your "riddle".

Btw. If pub = {}, then

ALL(a):[a in pub => a in drinkers]

holds (again vacuously) and hence, say,

{} in drinkers => ALL(a):[a in pub => a in drinkers]

holds (trivially). This implies

EXIST(x):[x in drinkers => ALL(a):[a in pub => a in drinkers]]

See: https://en.wikipedia.org/wiki/Vacuous_truth
and: https://en.wikipedia.org/wiki/Triviality_(mathematics)
and: https://en.wikipedia.org/wiki/Paradoxes_of_material_implication

[Chris M. Thomasson]

Reminds me of the Liar. Are we 100% sure a designated driver did not
sneak a couple of drinks in? I am 100% sober, I swear! Humm...

[Archimedes Plutonium]

On Wednesday, October 5, 2022 at 10:29:07 PM UTC-5, Chris M. Thomasson wrote:
> Try to tone down the anger a bit? A circle is a special case of an
> ellipse, right? For some reason, I think of an ellipse as a unit circle
> rotated in 3d.

Chris opened mouth before screwing in a brain in his head and comes up with-- a cat is a special case of a dog. A raspberry is a special case of a cherry. A square is a special case of a trapezoid. A rectangle is a special case of a square. A female is a special case of a male. The Sun is a special case of a planet. A airplane is a special case of a robin.

Why is this goonclod who has no degree in mathematics, why is this goonclod Chris posting daily to sci.math for 3 decades, when he should not be posting at all.

[Fritz Feldhase]

On Thursday, October 6, 2022 at 5:51:29 AM UTC+2, Archimedes Plutonium wrote:

> A square is a special case of a trapezoid.

Right, Archie!

See: https://en.wikipedia.org/wiki/Trapezoid#/media/File:Trapezoid_special_cases.png

[Archimedes Plutonium]

Fritz, the professional spammer, never studied Logic, and opened his mouth once again before screwing in a brain to his torso. How many Chris Thomassons does it take to screw in a brain for Fritz? Three, one to find the brain stem and hold it and one to screw in the brain. And one to give the instructions.

Fritz never studied logic or failed it, for there is a Fallacy that Fritz uses in abundance-- Irrelevancy. A grain of sand on a beach is a special case of Earth, but only a arsehole like Fritz wants to make it a issue.

[Chris M. Thomasson]

Wow! Unhinged. Do you think a circle is a special case of an ellipse
where the x and y radii are equal to one another?

[FromTheRafters]

Chris M. Thomasson formulated on Wednesday :

"The oval is not a precisely defined figure in mathematics."

"Ellipses are conic sections with eccentricity (e) between 0 and 1
while ovals are not precisely defined geometrical figures in
mathematics."

This is an oval.

https://www.beachcombingmagazine.com/blogs/news/a-mermaid-s-purse

[Ben Bacarisse]

The "paradox" is usually stated for non-empty pubs, and the person who
"exists such that..." is taken to be in the pub in question. Dan's
version is different in that the 'x' is not (necessarily) in the pub (or
even capable of being a drinker).

But, as is so often the case, Dan misses a chance to explain. This is
only a paradox of ambiguous language. The English wording contains two
elements of misdirection. Both the "if ... then ..." and the "there
exists ... such that" suggest causality which is not there in the logic.

If we replace the implication with a disjunction, thing are much
clearer: There is a person (in the pub) such that either (a) they are
not drinking, or (b) everyone is drinking. But even now the "such that
everyone is drinking" sounds as if this state of affairs might be being
"caused".

To avoid this the "there exists x such that..." is more clearly
expressed as choice: in any non-empty pub, we can find someone who is
either not drinking or is drinking with everyone else.

--
Ben.

[Mostowski Collapse]

Thats how you deal with your sci.logic and sci.math
community, pulling a dubious philosophy card?

LMAO!

[Mostowski Collapse]

I rather have the feeling your concept of "critical parameter" is
from some philosophy class. Maybe a philosophy class that
you did attend which was about "critical thinking?".

https://en.wikipedia.org/wiki/Critical_thinking

Or is the word "critical" from physics:

https://en.wikipedia.org/wiki/Critical_point_%28thermodynamics%29

Maybe its the later, since your "critical parameters" so far
made only DC Spooky explode, your alternative definitions
didn't yield anything good, at least not as you

advertised them.

[Dan Christensen]

The x needs to be a drinker, but not necessarily in the pub in question or in any other pub.

> But, as is so often the case, Dan misses a chance to explain. This is
> only a paradox of ambiguous language. The English wording contains two
> elements of misdirection. Both the "if ... then ..." and the "there
> exists ... such that" suggest causality which is not there in the logic.
>
> If we replace the implication with a disjunction, thing are much
> clearer: There is a person (in the pub) such that either (a) they are
> not drinking, or (b) everyone is drinking. But even now the "such that
> everyone is drinking" sounds as if this state of affairs might be being
> "caused".
>

[snip]

The key and most interesting part to me is that we have what seems to be a set theoretic variant of the principle of explosion or vacuous truth. By the basic rules of logic and set theory, we can prove that for ANY set S and proposition Q (be it true or false), we have:

EXIST(x):[x in S => Q]

In the case of the Drinker's Paradox, S is the set of all drinkers, and Q is the proposition that everyone in a given pub is drinking.

So, in "real world" scenarios, you may want to avoid having existential quantifiers on an implication. At the very least, it should set off warning bells and that you may want to consider a conjunction instead.

[Mostowski Collapse]

Except that your formalization is wrong. On the web you have:

EXIST(x):[x e drinkers => ALL(a):[a e pub => a e drinkers]]
http://www.dcproof.com/DrinkersThm1.htm

Shouldn't this be?

EXIST(x):[x e pub & [x e drinkers => ALL(a):[a e pub => a e drinkers]]]
https://en.wikipedia.org/wiki/Drinker_paradox

Which has a counter model, when pub can be empty:

∃x(px ∧ (dx → ∀a(pa → da))) is invalid.
https://www.umsu.de/trees/#~7x%28p%28x%29~1%28d%28x%29~5~6a%28p%28a%29~5d%28a%29%29%29%29

But otherwise is generally valid:

∃xpx → ∃x(px ∧ (dx → ∀a(pa → da))) is valid.
https://www.umsu.de/trees/#~7xp%28x%29~5~7x%28p%28x%29~1%28d%28x%29~5~6a%28p%28a%29~5d%28a%29%29%29%29

Nothing to do with Set Theory, Universal Set or Russel Paradox.
Since it is valid for classes, I guess it is also valid for sets.

[Mostowski Collapse]

You can use it to show that FOL assume a non-empty
universe of discourse, at least it is consistent with
the fact that this is also generally valid in FOL:

∃x(dx → ∀ada) is valid.
https://www.umsu.de/trees/#~7x%28d%28x%29~5~6ad%28a%29%29

It has also a dual:

∀z(dz → ∃xdx) → ∀z∃x(dz → dx) is valid.
https://www.umsu.de/trees/#~6z%28dz~5~7xdx%29~5~6z~7x%28dz~5dx%29

The two above possibly not provable in DC Spooky, right?

LoL

See also:

The Drinker Paradox and its Dual - 2018
Warren L, Diener H and McKubre-Jordens M
https://ir.canterbury.ac.nz/bitstream/handle/10092/16234/1805.06216v1.pdf

[Dan Christensen]

See my reply just now to your identical posting at sci.logic

Dan

On Thursday, October 6, 2022 at 11:22:03 AM UTC-4, Mostowski Collapse wrote:
> Except that your formalization is wrong. On the web you have:
>
> EXIST(x):[x e drinkers => ALL(a):[a e pub => a e drinkers]]
> http://www.dcproof.com/DrinkersThm1.htm
>
> Shouldn't this be?
>
> EXIST(x):[x e pub & [x e drinkers => ALL(a):[a e pub => a e drinkers]]]
> https://en.wikipedia.org/wiki/Drinker_paradox

[snip]

[Mostowski Collapse]

I wouldn't know that x e pub is redundant. The tree tool tells me
that it is not redundant. I get, for the crucial fact:

∃x(px ∧ (dx → ∀a(pa → da))) is invalid.
https://www.umsu.de/trees/#~7x%28p%28x%29~1%28d%28x%29~5~6a%28p%28a%29~5d%28a%29%29%29%29

But then:

∃x(dx → ∀a(pa → da)) is valid.
https://www.umsu.de/trees/#~7x(d(x)~5~6a(p(a)~5d(a)))

So nyet, x e pub is not redundant. If you think so, then there is
a bug somewhere in your blog. Or the problem is that

DC Spooky isn't FOL.

Dan Christensen schrieb am Donnerstag, 6. Oktober 2022 um 18:17:47 UTC+2:
> "x e pub" is redundant. The logic works for ANY drinker, whether in the pub or not. Maybe this was historically a feature of the original non-set-theoretical presentation of DP. The logic will also work for an empty pub.

[Cesario De rege]

Mostowski Collapse wrote:

> I wouldn't know that x e pub is redundant. The tree tool tells me that
> it is not redundant. I get, for the crucial fact:

you dirty bad teeth capitalists, there guys are not knowing what a NOT
capitalist is. They wanted kill you with *vaccines*, you fucking idiot,
because you are *too many*. This is literary, not a way a saying it.
*Literary*. How stoopid NOT capitalist can you be.

they made tests in the fucking nazi uKraine, in about 45 *bio_weapons*
illegal laboratories. Do you need more proofs, you fucking nazis.

Outraged UK Citizens ARREST CopsFor Failing To Shut Down Nuremberg-
Violating mRNA Shot Center https://www.bitchute.com/video/YbUtwerdPXEm/

the following means the *comradeship* between the *working_people* in
*industry* with the working people in *agriculture*.

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[Dan Christensen]

On Wednesday, October 5, 2022 at 4:12:23 PM UTC-4, Dan Christensen wrote:
> The Drinker’s Theorem: Consider the set of all drinkers in the world, and the set of all people in a given pub. Then there exists a person who, if he or she is drinking, then everyone in that pub is drinking.
>

> It doesn’t matter how many people are there. Or how many are drinking. Or how few. No one needs to be taking their cues from some “lead drinker,” but in every pub, in every town and village, it just happens! How is this possible?
>

> There are several possible approaches to this problem. Here, we will turn to British philosopher and mathematician, Bertrand Russell (1872 – 1970). His famous Paradox is the key.
>

I don't know if this is a truly novel approach to DP, but, on Google right now, after 8 years, my blog posting still seems to be the only online connection between the two paradoxes.

From Wikipedia: "The drinker paradox (also known as the drinker's theorem, the drinker's principle, or the drinking principle) is a theorem of classical predicate logic that can be stated as "There is someone in the pub such that, if they are drinking, then everyone in the pub is drinking."
https://en.wikipedia.org/wiki/Drinker_paradox

My version of DP is based on set theory. It does not require the "lead drinker" to be in the pub. In fact, anything whatsoever is implied by his/her very existence! How cool is that???

It is a theorem of basic set theory that for any set S and ANY proposition Q (be it true or false), we have EXIST(x):[x in S => Q]. See....

> See: https://dcproof.wordpress.com/2014/06/03/the-drinkers-paradox/

[Dan Christensen]

On Thursday, October 6, 2022 at 3:57:53 PM UTC-4, Mostowski Collapse wrote:
> I wouldn't know that x e pub is redundant. The tree tool tells me
> that it is not redundant. I get, for the crucial fact:
> ∃x(px ∧ (dx → ∀a(pa → da))) is invalid.
> https://www.umsu.de/trees/#~7x%28p%28x%29~1%28d%28x%29~5~6a%28p%28a%29~5d%28a%29%29%29%29
> But then:
>
> ∃x(dx → ∀a(pa → da)) is valid.
> https://www.umsu.de/trees/#~7x(d(x)~5~6a(p(a)~5d(a)))
>
> So nyet, x e pub is not redundant.

It may not be redundant in the original predicate logic version, but it is certainly redundant in my set theoretic version.

> If you think so, then there is
> a bug somewhere in your blog.

See if you can find it! (Don't hold your breath, folks.)

[Earle Jones]

On Wed Oct 5 16:34:06 2022 Dan Christensen wrote:

> On Wednesday, October 5, 2022 at 5:24:12 PM UTC-4, Johnnie Bertolini wrote:
> > Dan Christensen wrote:
> >
> > > The Drinker s Theorem: Consider the set of all drinkers in the world,
> > > and the set of all people in a given pub. Then there exists a person
> > > who, if he or she is drinking, then everyone in that pub is drinking.
>

> > AFU Unit Describes Command Ordering Them To Shoot Every Civilian In Town
> > And Their Own Wounded
>
> And you support this, Nazi boy???
>
> So, have you been, ahem... "volunteered" to die for Putin's massive ego yet? It may be too late to flee Nazi ruSSia. Maybe go into hiding for a few years in a remote little dacha in the forest in the far north. It might be a nice change for you. Be careful who you talk to though. You never know who might turn you in.

*
DAN: Your mathematics is great. But your political skills (like mine) leave somethng to be desired.

Let's stay our of politics and into mathematics.

Cheers!!

earle
*

[Ross A. Finlayson]

Weebles, ... wobble: but don't fall down.

[Dan Christensen]

Politics???

[Mostowski Collapse]

Nope, its also not redundant set theoretically.
Take Exy for x e y:

∃x(Exp ∧ (Exd → ∀a(Eap → Ead))) is invalid.
https://www.umsu.de/trees/#~7x%28Exp~1%28Exd~5~6a%28Eap~5Ead%29%29%29

∃x(Exd → ∀a(Eap → Ead)) is valid.
https://www.umsu.de/trees/#~7x%28Exd~5~6a%28Eap~5Ead%29%29

[Mostowski Collapse]

The problem is you don't show the drinker paradox here:
You only prove:

EXIST(x):[x e drinkers => ALL(a):[a e pub => a e drinkers]]
http://www.dcproof.com/DrinkersThm1.htm

But you should show that this here is not provable,
but unlike the tree tool, DC Spoiled is not a tool to show counter models.

EXIST(x):[x e pub & [x e drinkers => ALL(a):[a e pub => a e drinkers]]]

And that this here is provable, this is something you could
at least do with DC Spoild, not 100% sure.

EXIST(x):[x e pub] => EXIST(x):[x e pub & [x e drinkers => ALL(a):[a e pub => a e drinkers]]]

[Mostowski Collapse]

One really wonders what you were smoking when you
did the Drinker paradox. It doesn't follow Occams razor,
its just some backed nonsense. You use some withess x

such that ~(x e drinkers) to show. Which you first draw
from Rusells paradox it seems:

EXIST(x):[x e drinkers => ALL(a):[a e pub => a e drinkers]]
http://www.dcproof.com/DrinkersThm1.htm

But the proof here:

∃x(Exd → ∀a(Eap → Ead)) is valid.
https://www.umsu.de/trees/#~7x%28Exd~5~6a%28Eap~5Ead%29%29

Proceeds the same way like the proof here, the second a e pub is redundant:

∃x(Exd → ∀aEad) is valid.
https://www.umsu.de/trees/#~7x%28Exd~5~6aEad%29

Both proofs succeeds because they lead to:
~Ebd , Ebd

Noting to do with really deriving ~(x e drinkers) first from
some Russel paradox as you do. Thats utter nonsense. The drinker
paradox is LEM together with some quantifier rules.

p v ~p
https://en.wikipedia.org/wiki/Law_of_excluded_middle

[Mostowski Collapse]

Interestingly both proofs use also contraction.
But they don't need Russells paradox per se.
The original statement is copied twice.

Although some versions of Russells paradox
and also some versions of Currys paradox, also
depend on contraction. So maybe you have a small

point. But Russells paradox uses x e x, which I
don't see in the Drinker paradox. Is there somewhere
drinker e drinker or x e x or pub e pub? No, not at all!

What an utter nonsense!

[Mostowski Collapse]

Here is a proof of the Drinker paradox, that doesn't
use first Russells Paradox. Namely I find speedily,
even in DC Spoiled:

21 ALL(d):EXIST(x):[x ε d => ALL(a):a ε d]
Rem DNeg, 20

But it has a twist, it indeed leads to some Russells
like d e d (sic!). But this happens out of sheer lack
to invent new variables during a proof,

other proof tools don't need this trick. If I have time
I will show such a proof tool. I wasn't successfull with
the Wolfgang Schwartz tree tool, it also falls back to

d e d, see this GitHub issue:

Strange tendency to turn the Drinker Paradox into a Russell Paradox
https://github.com/wo/tpg/issues/21

One can also look at the proofs here, none of the
proofs does some Russell stuff like for example P(P),
which would be the analogue if you are not in set theory,

so maybe Dan Christensen has become slave of his tool?
Not master of logic anymore, but slave of what the tool
dictates, including some Russell nonsense?

Proof of Drinker paradox [duplicate]
https://math.stackexchange.com/questions/807092/proof-of-drinker-paradox

--------------------------- begin proof ---------------------

1 ~EXIST(x):[x ε d => ALL(a):a ε d]
Premise

2 ~~ALL(x):~[x ε d => ALL(a):a ε d]
Quant, 1

3 ALL(x):~[x ε d => ALL(a):a ε d]
Rem DNeg, 2

4 ~[d ε d => ALL(a):a ε d]
U Spec, 3

5 ~~[d ε d & ~ALL(a):a ε d]
Imply-And, 4

6 d ε d & ~ALL(a):a ε d
Rem DNeg, 5

7 d ε d
Split, 6

8 ~ALL(a):a ε d
Split, 6

9 ~~EXIST(a):~a ε d
Quant, 8

10 EXIST(a):~a ε d
Rem DNeg, 9

11 ~b ε d
E Spec, 10

12 ~[b ε d => ALL(a):a ε d]
U Spec, 3

13 ~~[b ε d & ~ALL(a):a ε d]
Imply-And, 12

14 b ε d & ~ALL(a):a ε d
Rem DNeg, 13

15 b ε d
Split, 14

16 ~ALL(a):a ε d
Split, 14

17 ~b ε d & b ε d
Join, 11, 15

18 ~EXIST(d):~EXIST(x):[x ε d => ALL(a):a ε d]
Conclusion, 1

19 ~~ALL(d):~~EXIST(x):[x ε d => ALL(a):a ε d]
Quant, 18

20 ALL(d):~~EXIST(x):[x ε d => ALL(a):a ε d]
Rem DNeg, 19

21 ALL(d):EXIST(x):[x ε d => ALL(a):a ε d]
Rem DNeg, 20

--------------------------- end proof ---------------------

[Phil Carmody]

Ben Bacarisse <ben.u...@bsb.me.uk> writes:
> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>
>> On 10/5/2022 7:24 PM, Fritz Feldhase wrote:
>>> On Thursday, October 6, 2022 at 3:57:16 AM UTC+2, Chris M. Thomasson wrote:
>>>> On 10/5/2022 1:12 PM, Dan Christensen wrote:
>>>>>
>>>>> The Drinker’s Theorem: Consider the set of all drinkers in the
>>>>> world, and the set of all people in a given pub. Then there
>>>>> exists a person who, if he or she is drinking, then everyone in
>>>>> that pub is drinking.

...

> But, as is so often the case, Dan misses a chance to explain. This is
> only a paradox of ambiguous language. The English wording contains two
> elements of misdirection. Both the "if ... then ..." and the "there
> exists ... such that" suggest causality which is not there in the logic.
>
> If we replace the implication with a disjunction, thing are much
> clearer: There is a person (in the pub) such that either (a) they are
> not drinking, or (b) everyone is drinking. But even now the "such that
> everyone is drinking" sounds as if this state of affairs might be being
> "caused".
>
> To avoid this the "there exists x such that..." is more clearly
> expressed as choice: in any non-empty pub, we can find someone who is
> either not drinking or is drinking with everyone else.

From a language perspective, I'd say that these are still worded so that
they are false. The "is drinking" is a continuing action, and the "is"
implies constancy, thus continuing to be true while the "is drinking"
applies. The is_drinking predicate D() is not constant, it's varying
over time - and the drinker, likewise, needs to be able to vary over
time. So there doesn't "exist *a* drinker", as that implies constancy,
which is inappropriate.

The paradox is is worded more like:
exists i in P : forall t D(i,t) -> (forall j in P D(j,t))
but the logical explanation is explaining this:
forall t exists i(t) in P : D(i(t),t) -> (forall j in P D(j,t))

That's not even a swap of the quantifiers, that's a more significant
change.

Word things badly, in ways that overlook how language is interpreted,
and you'll get contradictory conclusions. Them's the breaks.

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have
gained some understanding of the world in which we live. As such, we can cast
aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/

[FromTheRafters]

Ross A. Finlayson has brought this to us :

I saw a butterfly lay an egg on a leaf. I photographed it and tried to
identify the species by the photo of the egg. As it turns out, it was
an Eastern Tiger Swallowtail.

While reading about the ETS I noticed they described the egg's features
and stated that it was oval in shape. So, I looked up oval to find out
that it simply means 'egg shaped' -- who knew, it apparently lays egg
shaped eggs.

[Dan Christensen]

On Friday, October 7, 2022 at 3:28:15 AM UTC-4, Mostowski Collapse wrote:
> Here is a proof of the Drinker paradox, that doesn't
> use first Russells Paradox.

[snip]

Well done, Jan Burse! Line 4 with "d in d" bothers me a bit, but it works.

Now prove the more general case, given any proposition Q and set s to obtain EXIST(b):[b in s => Q]. I think you will need RP for that (see my proof).

Then DP will simply be a special case of this mind-blowing little theorem.

[Ross A. Finlayson]

We had a thread about Curry paradox, I broke it up.

[Mostowski Collapse]

The proof looks different if you use this:
Naoyuki Tamura seqprover for classical first-order logic.

------------- Ax
d(Z) --> d(Z)
--------------- R#
d(Z) --> Y#d(Y)
----------------- R->
--> d(Z)->Y#d(Y)
--------------------- R@
--> X@(d(X)->Y#d(Y))
------------------------ Ltop
top --> X@(d(X)->Y#d(Y))
https://www.vidal-rosset.net/g4-prover/g4action.php?formula=top+--%3E+X%40%28d%28X%29-%3EY%23d%28Y%29%29&sysopt=g4c&output=pretty

It invents a new variable Z, no need for d(d).
Thats the normal way to prove the Drinker paradox,

there is something strange going on with
Wolfgrang Schwartz tree tool.

[Mostowski Collapse]

The proof is also contraction free, and therefore shorter.
The original formula is only used once, and not twice.
In DC Spoiled I didn't manage to use the original formula

only once, I made the negated original formula a premis
to carry out the proof. But I had to use the original formula
twice. Is there a DC Spoiled proof which is as short as

the sequent calculus proof?

[Ross A. Finlayson]

"Curry's poor substitute" is what you got there.

"Propagation of the liar paradox", quit pushing it.

"Template of wrong".

Curry's Paradox in propositional logic?



That's non-monotonic.



Yeah, that Curry thread has lots of great quotes.

Helps break down five or six paradoxes in one thread.

[Mostowski Collapse]

Oops, I guess I confused ALL() and EXISTS(),
but still no Russell Paradox, which would read d(d):

------------------------------------------- Ax-c
d(X1),d(Z) --> d(X1),d(Y1),X#(d(X)->Y@d(Y))
-------------------------------------------- R@
d(X1),d(Z) --> d(X1),Y@d(Y),X#(d(X)->Y@d(Y))
--------------------------------------------- R->
d(Z) --> d(X1),d(X1)->Y@d(Y),X#(d(X)->Y@d(Y))
--------------------------------------------- R#-c
d(Z) --> d(X1),X#(d(X)->Y@d(Y))
-------------------------------- R@
d(Z) --> Y@d(Y),X#(d(X)->Y@d(Y))
---------------------------------- R->
--> d(Z)->Y@d(Y),X#(d(X)->Y@d(Y))
---------------------------------- R#-c
--> X#(d(X)->Y@d(Y))
------------------------ Ltop
top --> X#(d(X)->Y@d(Y))

https://www.vidal-rosset.net/g4-prover/g4action.php?formula=top+--%3E+X%23%28d%28X%29-%3EY%40d%28Y%29%29+&sysopt=g4c&output=pretty

Contraction is still there!
And it invents a little more variables, Z, X1, Y1

[Dan Christensen]

On Friday, October 7, 2022 at 1:03:35 PM UTC-4, Mostowski Collapse wrote:
> Oops, I guess I confused ALL() and EXISTS(),
> but still no Russell Paradox, which would read d(d):
>
> ------------------------------------------- Ax-c
> d(X1),d(Z) --> d(X1),d(Y1),X#(d(X)->Y@d(Y))
> -------------------------------------------- R@
> d(X1),d(Z) --> d(X1),Y@d(Y),X#(d(X)->Y@d(Y))
> --------------------------------------------- R->
> d(Z) --> d(X1),d(X1)->Y@d(Y),X#(d(X)->Y@d(Y))
> --------------------------------------------- R#-c
> d(Z) --> d(X1),X#(d(X)->Y@d(Y))
> -------------------------------- R@
> d(Z) --> Y@d(Y),X#(d(X)->Y@d(Y))
> ---------------------------------- R->
> --> d(Z)->Y@d(Y),X#(d(X)->Y@d(Y))
> ---------------------------------- R#-c
> --> X#(d(X)->Y@d(Y))
> ------------------------ Ltop
> top --> X#(d(X)->Y@d(Y))
>

Have you been able to prove that, for any proposition Q, we have:

ALL(s):[Set(s) => EXIST(a):[a in s => Q]]?

Or ALL(s):EXIST(a):[a in s => Q]? (Without my Set predicate).

That's the really interesting part of all this. DP then becomes just a special case of this theorem.

[Ross A. Finlayson]

No, that's the difference between a "proof" and your instancy.

I think you might find some disagreement what the most, "interesting", part is.

What are you trying to prove?

[Ross A. Finlayson]

Contraction is the course, ....

[Dan Christensen]

For any proposition Q, we have:

ALL(s):[Set(s) => EXIST(x):[x in s => Q]]

[Mostowski Collapse]

Its not relevant to the drinker paradox. You are completely
on the wrong track about the drinker paradox. Who
led you on this nonsense track wonky man? Let me explain.

Obviously when you prove the below sequent calculus proof,
and the sequent calculus uses contraction:

------------------------ Ltop
top --> X#(d(X)->Y@d(Y))

Which is equivalent to proving:

-------------------------------------------
EXIST(x):[d(x) => ALL(y):d(y)]

The because the proof uses contraction, this is a sign
that there is some PROOF BY CASES going on. There
are these two cases namely:

Case 1: EXIST(x):~d(x), or lets say ~d(x0) and x0
is the witness. We then have ex falso quodlibet,
we can choose for the x in EXIST(x) the value x0,

and then get:

~d(x0) => ALL(y):d(y)

Which is true since ~d(x0) true. Simple
material implication. It is row 1. and 2. here:

A B A => B
1. 0 0 1
2. 0 1 1
3. 1 0 0
4. 1 1 1

Case 2: ~EXIST(x):~d(x), this logically equivalent
to ALL(y):d(y). We then have verum stays verum. So
the contrary case to case 1, is case 2 where ALL(y):d(y)

is true and we get again, now irrelevant what z is:

~d(z) => ALL(y):d(y)

Which is true since ALL(y):d(y) is true. Simple
material implication. It is row 2. and 4. here:

A B A => B
1. 0 0 1
2. 0 1 1
3. 1 0 0
4. 1 1 1

Got it? Do you understand the Drinker Paradox now?

[Mostowski Collapse]

Your nonsense proof here is not the Drinker Paradox:
http://www.dcproof.com/DrinkersThm1.htm

It doesn't have contraction, i.e. this inference rule:

A v A
---------------------------- (Contr)
A

which would be a sign that the formula A is needed from
two different angles. You start with some nonsense

EXIST(a):~a e drinkers
Detach, 4, 2

Which is only case 1, but there are two cases,

Case 1: EXIST(x):~d(x)

Case 2: ~EXIST(x):~d(x)

Both make the Drinker Paradox formula true.
Since they are opposite cases, you can use proof
by cases as a proof method, basically this here:

(A => B) & (~B => A) => A

How old is your Drinker Pradox? Damn wonky
man, do you want to tell us you wasted 18 years
also here, its not only the function spaces,

but also the Drinker Paradox which challenges
you already for years and years? Congratulations
you are a true crank, much worse than AP brain farto.

[Mostowski Collapse]

I doubt that your set theoretic take is still the drinker
paradox. What is the ambivalence of the drinker paradox,
if you have anyway a non-drinker? There is no ambivalence

anymore, there is no paradoxical situation, in that contrary
situations lead to the same outcome. You can go about
the internet, and easily find the approach to "resolve" the

drinker paradox, in that one proves this ambivalence, by
proof by cases, in pratically all corners of the internet,
like for example here, the case 2 and case 1 explaind:

"Proof of the paradox
The proof begins by recognising it is true that either everyone
in the pub is drinking (in this particular round of drinks), or
at least one person in the pub isn't drinking."
https://paradox.fandom.com/wiki/Drinker_paradox

[Mostowski Collapse]

You can then go further, for experts only, and draw
from it further didactic value, quite different from
your marriage in hell between Russell and Smullyan,

"This proof illustrates several properties of classical
predicate logic which do not always agree with
ordinary language.
Non-Empty Domain
Excluded Middle"
https://paradox.fandom.com/wiki/Drinker_paradox

[Urbano Stilo]

Mostowski Collapse wrote:

> You can then go further, for experts only, and draw from it further
> didactic value, quite different from your marriage in hell between
> Russell and Smullyan,
>
> "This proof illustrates several properties of classical predicate logic
> which do not always agree with ordinary language.
> Non-Empty Domain Excluded Middle"

you can't even fart for hunger.

[Urbano Stilo]

Dan Christensen wrote:

> On Friday, October 7, 2022 at 3:28:15 AM UTC-4, Mostowski Collapse
> wrote:
>> 20 ALL(d):~~EXIST(x):[x ε d => ALL(a):a ε d] Rem DNeg, 19 21
>> ALL(d):EXIST(x):[x ε d => ALL(a):a ε d] Rem DNeg, 20
>
> Well done, Jan Burse! Line 4 with "d in d" bothers me a bit, but it
> works. Now prove the more general case, given any proposition Q and set s
> to obtain EXIST(b):[b in s => Q]. I think you will need RP for that (see
> my proof).

we fart you gently.

[Dan Christensen]

> Its not relevant to the drinker paradox.

[snip]

Wrong again, Jan Burse. If you let s be the set of all drinkers and Q be the proposition that everyone in the pub in question is a drinker, then you get the required result (requires RP AFAIK). Get it? Didn't think so. Oh, well...

[Mostowski Collapse]

Show us a DC Spoiled proof of your nonsense, and we will
be able to pinpoint more clearly what goes wrong. Its always
the same game, circle of life of DC Spoiled, a lot of bla bla and

then some retract. It is not the desired result. You dont understand
paradoxes, and if they use phrases such as:

- Drinker Paradox: The people in a pub ...
- Liar Paradox: The people on an island ...
- Etc... Etc...

You don't understand what this means right? Thats only another
way to say domain of discourse. You showed yourself that
there is no universal set, but if pub stands for the domain of discourse,

how can you say for example this here:

EXIST(x):[x e drinkers => ALL(a):[a e pub => a e drinkers]]
http://www.dcproof.com/DrinkersThm1.htm

Its utter nonsense. You turn the domain of discourse pub into a set.
That will not work, its already the wrong modelling. On the other hand
a predicate P(_) could have the extension of the domain of discourse,

put _ e pub cannot have the extension of the domain of discourse. Here
is a proof that pub cannot have the extension of the domain of discourse,
using your own previouse result:

/* Previous Result http://www.dcproof.com/UniversalSet.htm */
1 ALL(s):[Set(s) => EXIST(a):~a ε s]
Axiom

2 Set(pub)
Axiom

3 Set(pub) => EXIST(a):~a ε pub
U Spec, 1

4 EXIST(a):~a ε pub
Detach, 3, 2

[Dan Christensen]

On Friday, October 7, 2022 at 7:37:55 PM UTC-4, Mostowski Collapse wrote:
> Show us a proof [snip childish abuse], and we will

> be able to pinpoint more clearly what goes wrong.

[snip nonsense]

Here it is again: http://www.dcproof.com/STGeneralizedDrinkersThm.htm

Read it and weep, Jan Burse. Where do imagine I "go wrong?" (Don't hold your breath, folks!)

[Urbano Stilo]

Dan Christensen wrote:

> On Friday, October 7, 2022 at 7:37:55 PM UTC-4, Mostowski Collapse
> wrote:
>> Show us a proof [snip childish abuse], and we will be able to pinpoint
>> more clearly what goes wrong.
>
> [snip nonsense]

this gay actor cocaine khazar satanist is unwashed. They made him
"president". Proofs:

Zelensky： A Day of Infamy, Ignored by Mainstream Press
https://www.bitchute.com/video/y4vM3kNM7Rwy/

[Dan Christensen]

On Saturday, October 8, 2022 at 12:43:08 AM UTC-4, Urbano Stilo wrote:
> Dan Christensen wrote:
>
> > On Friday, October 7, 2022 at 7:37:55 PM UTC-4, Mostowski Collapse
> > wrote:
> >> Show us a proof [snip childish abuse], and we will be able to pinpoint
> >> more clearly what goes wrong.
> >
> > [snip nonsense]
> this gay actor cocaine khazar satanist is unwashed. They made him
> "president". Proofs:
>
> Zelensky： A Day of Infamy, Ignored by Mainstream Press

So far, it's 227 days of infamy for your boss, the war criminal Putin. And you will be sitting next to him in the Nuremberg Tribunal II, Nazi boy. He will probably try to blame you, for feeding him all this fake news. If you can't flee the country, you had better go into hiding. This will not end well for you.

[Urbano Stilo]

Dan Christensen wrote:

> On Saturday, October 8, 2022 at 12:43:08 AM UTC-4, Urbano Stilo wrote:
>> Dan Christensen wrote:
>>
>> > On Friday, October 7, 2022 at 7:37:55 PM UTC-4, Mostowski Collapse
>> > wrote:
>> >> Show us a proof [snip childish abuse], and we will be able to
>> >> pinpoint more clearly what goes wrong.
>> >
>> > [snip nonsense]
>> this gay actor cocaine khazar satanist is unwashed. They made him
>> "president". Proofs:
>>
>> Zelensky： A Day of Infamy, Ignored by Mainstream Press
>
> So far, it's 227 days of infamy for your boss, the war criminal Putin.
> And

how "fake news" you are snipping the link to the context, when the
journalist is an american personality. You fool. Here we go, one more time.
I'm just printing the proofs. You wankers don't know what a proof is. You
are deleting the links to proofs. The judge will find you guilty for that
too. You are guilty like a dog.

Zelensky： A Day of Infamy, Ignored by Mainstream Press

https://www.bitchute.com/video/y4vM3kNM7Rwy/

[Dan Christensen]

On Saturday, October 8, 2022 at 1:06:11 AM UTC-4, Urbano Stilo wrote:
> Dan Christensen wrote:
>
> > On Saturday, October 8, 2022 at 12:43:08 AM UTC-4, Urbano Stilo wrote:
> >> Dan Christensen wrote:
> >>
> >> > On Friday, October 7, 2022 at 7:37:55 PM UTC-4, Mostowski Collapse
> >> > wrote:
> >> >> Show us a proof [snip childish abuse], and we will be able to
> >> >> pinpoint more clearly what goes wrong.
> >> >
> >> > [snip nonsense]
> >> this gay actor cocaine khazar satanist is unwashed. They made him
> >> "president". Proofs:
> >>
> >> Zelensky： A Day of Infamy, Ignored by Mainstream Press
> >
> > So far, it's 227 days of infamy for your boss, the war criminal Putin.

> > And you will be sitting next to him in the Nuremberg Tribunal II, Nazi boy. He will probably try to blame you, for feeding him all this fake news. If you can't flee the country, you had better go into hiding. This will not end well for you.

> how "fake news" you are snipping the link to the context..

Tell it to the judges at your trial, Nazi boy. Give them all your links to bitchute.com See how impressed they will be. XAXAXA If you can't ride, you had better hide, Nazi boy.

[Michael Moroney]

It's too late, nymshifter! I told you to get out while the getting was
good. Not only is the border closed everywhere, but you can't even get
to 卐Ru⚡︎⚡︎ian卐 Occupied Crimea any longer! It looks like they took out
the Kerch Bridge! What a wonderful birthday present for 卐Владольф
Путлер卐 and his 70th Birthday!

[Mostowski Collapse]

About Dan Christensens Schrödinger Drinking Paradox:

You produced generalized nonsense. You will be possibly remembered
as the most schizophrenic crank on sci.math and sci. logic. First
you start with this nonsense:
1) Sets are Bad 1: Sets are bad, because with ordinals we
would have 3 e 4, i.e. less than becomes membership.
2) Sets are Bad 2: Sets are bad, because as far as I (wonky man)
can tell, mathe text books don't use set theory.
3) What else from the mouth of Dan Christensen (wonky man)?

As soon as one has rubbed his eyes to see the above nonsense
more cleary, Dan Christensen turns around the corner and claims
the contrary, namely:
4) We need Sets 1: If you tell him we work in the natural numbers,
he turns around the corner yes, but you need to use x e N.
5) We need Sets 2: Although Drinker Paradox is a FOL paradox,
he now believes its a FOL+SET paradox.
6) What else from the mouth of Dan Christensen?

Whats next, the moon is made out of green cheese and
I (wonky man) am a Cat and not a Cat?

[Mostowski Collapse]

To the best of my knowledge the Drinker Paradox is a FOL
paradox and not a FOL+SET paradox, you can read here:

"Proof of the paradox
The proof begins by recognising it is true that either everyone
in the pub is drinking (in this particular round of drinks), or
at least one person in the pub isn't drinking."
https://paradox.fandom.com/wiki/Drinker_paradox

And if you would dig deeper, it would even reveal some
miractles about the below topics, I have highlighted the
phrase **classical predicate logic** aka FOL.

"This proof illustrates several properties of **classical
predicate logic** which do not always agree with

ordinary language.
Non-Empty Domain
Excluded Middle"

https://paradox.fandom.com/wiki/Drinker_paradox

The term FOL alone doesn't imply that we look at FOL+SET.
But you need to leave classical predicate logic for a deeper
understanding, which you are completely lacking, since you

look at FOL+SET. On the quantifier side this could be free
logic (for the Non-Empty Domain issue). But of course non-
classicality in the propositional connectives might also be

interesting to dig deeper (for the Excluded Middle issue).
How the later is done, you might want to consult this here:

The Drinker Paradox and its Dual - Louis Warren et al., 2018
https://ir.canterbury.ac.nz/bitstream/handle/10092/16234/1805.06216v1.pdf

[Dan Christensen]

On Saturday, October 8, 2022 at 11:18:38 AM UTC-4, Mostowski Collapse wrote:
> To the best of my knowledge the Drinker Paradox is a FOL
> paradox and not a FOL+SET paradox, you can read here:

Pay attention, Jan Burse. Recall that I wrote in my original posting here: "There are several possible approaches to this problem. Here, we will turn to British philosopher and mathematician, Bertrand Russell (1872 – 1970). His famous Paradox is the key."

AFAICT this is an original approach. It makes use of the theorem: For any set S and proposition Q, we have EXIST(x):[x in S => Q].

[Fritz Feldhase]

On Saturday, October 8, 2022 at 6:31:25 PM UTC+2, Dan Christensen wrote:
>
> It makes use of the theorem: For any set S and proposition Q, we have EXIST(x):[x in S => Q].

This certainly will hold in the context of ZF(C). But there are set theories with an "universal set", say V, with Ax(x e V).

There ~Ex(x e V -> x =/= x) (for example) is a theorem.

[Dan Christensen]

On Saturday, October 8, 2022 at 12:52:14 PM UTC-4, Fritz Feldhase wrote:
> On Saturday, October 8, 2022 at 6:31:25 PM UTC+2, Dan Christensen wrote:
> >
> > It makes use of the theorem: For any set S and proposition Q, we have EXIST(x):[x in S => Q].

> This certainly will hold in the context of ZF(C). But there are set theories with an "universal set", say V, with Ax(x e V).
>

The only axiom of set theory I used to prove the non-existence of the set of all things (the supposed "universal" set) is the Subset Axiom ("Specification" in ZFC). To have a universal set, it seems to me that you would have to somehow restrict the application of that axiom, or toss it altogether. Neither sounds like a very appealing alternative.

And yet, from Wikipedia:

"In set theory and its applications throughout mathematics, a class is a collection of sets (or sometimes other mathematical objects) that can be unambiguously defined by a property that all its members share. Classes act as a way to have set-like collections while differing from sets so as to avoid Russell's paradox (see § Paradoxes). The precise definition of "class" depends on foundational context. In work on Zermelo–Fraenkel set theory, the notion of class is informal [a bit too hand-wavy for my liking--DC], whereas other set theories, such as von Neumann–Bernays–Gödel set theory, axiomatize the notion of "proper class", e.g., as entities that are not members of another entity."
https://en.wikipedia.org/wiki/Class_(set_theory)

As it is, ZFC is rarely mentioned in textbooks. NBG even more so. I think I will pass on that one, too. Thanks anyway.

[Michel Marconi]

Dan Christensen wrote:

> On Saturday, October 8, 2022 at 11:18:38 AM UTC-4, Mostowski Collapse
> wrote:
>> To the best of my knowledge the Drinker Paradox is a FOL paradox and
>> not a FOL+SET paradox, you can read here:
>
> Pay attention, Jan Burse. Recall that I wrote in my original posting
> here:
> "There are several possible approaches to this problem. Here, we will
> turn

you nazis are guilty for genocide in the eastern "uKraine", a fictitious
nazi state. Most likely you nazis will be judged by the people of the
*Popular_Republic_of_Donesk* and the *Popular_Republic_of_Lugansk* then
executed by hanging in a tree.

[Dan Christensen]

I'm not worried about that possibility, but you should be, Nazi boy.

[Mostowski Collapse]

Bertrand Russell never approached the Drinker Paradox.
You are crazy. His Russell Paradox is from 1901, at
that time classical predicate logic wasn't even completely

born. On the other hand the Drinker Paradox is from
1978 and from Raymond Smullyan, where people already
knew more about classical predicate logic, what we

nowadays call FOL. So how should have Russell
have adressed Smullyan. Did Time flow backwards?

You are a nuthead!

[Mostowski Collapse]

The current FOL is from Frege and Gödel. For example
Gödel clearly explains how predicate logic can properly
deal with empty domains, whereas the Aristotelean

Logic Lectures: Gödel's Basic Logic Course at Notre Dame
https://arxiv.org/abs/1705.02601

Term Logic often had existential import. The current
FOL setting is from around 1950s after Bertrand Russell.
That FOL is possible to cover both empty domains

and non empty domains is easily seen by this switchero.

Empty Domains Off:

/* Provable in FOL */
∃x(D(x) => ∀yD(y))

Empty Domains On:

/* Not Provable in FOL */
∃x(P(x) & (D(x) => ∀y(P(y) => D(y)))

The problem is that your DC Spoiled is also some nonsense
from before 1950. You have even botched it over the time
more, since for example you have tweeked some quantifier

rules, and what horror connected it to sets! What a complete
utter nonsense. Your DC Spoiled can even not handle FOL
properly. And your conflation of Russel and Smullyan has

the only goal to write some blog, although DC Spoiled
has nothing to contribute to the Drinker Paradox, since it
is not properly FOL, it can even not prove:

/* Not Provable in DC Spoiled, but Provable in FOL */
∃x(D(x) => ∀xD(x))

[Dan Christensen]

On Saturday, October 8, 2022 at 4:33:55 PM UTC-4, Mostowski Collapse wrote:
> Bertrand Russell never approached the Drinker Paradox.
> You are crazy. His Russell Paradox is from 1901, at
> that time classical predicate logic wasn't even completely
>
> born. On the other hand the Drinker Paradox is from
> 1978 and from Raymond Smullyan, where people already
> knew more about classical predicate logic, what we
>
> nowadays call FOL. So how should have Russell
> have adressed Smullyan.

What do you mean? I saw them both just the other day! Or was it 50 years ago? Anyway, the 3 of us had a beer together in Prague. Bertie bought a round of Guinness for everyone in the pub. When Bertie drinks, EVERYONE drinks. Cheers. Ein prosit!

Dan

[Mostowski Collapse]

Dang wonky man. You got it wrong!

When Bertie doesn't drink, then there is somebody
whos drinking implies that everyone drinks.
Thats case 2 here:

"proof begins by recognising it is true that either everyone
in the pub is drinking (in this particular round of drinks), or
at least one person in the pub isn't drinking."
https://paradox.fandom.com/wiki/Drinker_paradox

What do you want to deduce from Bertie drinks?
Tell us wonky man?

Also please show us in DC Spoiled this theorem:

/* Not Provable in DC Spoiled, but Provable in FOL */

∃x(D(x) => ∀yD(y))

[Mostowski Collapse]

Paying a round of Guinness for everyone in the pub
doesn't imply that everybody drinks.

[Fritz Feldhase]

On Saturday, October 8, 2022 at 8:27:46 PM UTC+2, Dan Christensen wrote:
> On Saturday, October 8, 2022 at 12:52:14 PM UTC-4, Fritz Feldhase wrote:
> > On Saturday, October 8, 2022 at 6:31:25 PM UTC+2, Dan Christensen wrote:
> > >
> > > It makes use of the theorem: For any set S and proposition Q, we have EXIST(x):[x in S => Q].
> > >
> > This certainly will hold in the context of ZF(C). But there are set theories with an "universal set", say V, with Ax(x e V).
> >

> Neumann–Bernays–Gödel set theory

No, there we have a "universal class" which contains just all sets. But since V is a proper class (and hence no set) still ~ (V e V).

This means that your theorem holds at least in ZF(C) and NBG.

Though it has nothing to do with the "Drinker Paradox".

[Mostowski Collapse]

Dan Christensen is the uber-crank. He not only beats John Gabriel,
he is on par with "John Garbiel: look I reinvented calculus, here is
new calculus", in that he has "Dan Christensen: look I reinvented

logic, here is DC Proof". He also beats Archimedes Plutonium,
in that he has more Alzheimer than Archimedes Plution. You
can talk to Dan Christensen and he doesn't understand acronyms:

FOL = first order logic
FOL+SET = first order logic plus his Subset axiom schema

And he doesn't know facts like:

Drinker Paradox is a FOL Paradox
Drinker Paradox is not a FOL+SET Paradox
Russell Paradox is not a FOL Paradox
Russell Paradox is a FOL+SET Paradox

Please wonky many, show as the Driker Paradox in FOL
only? What is FOL only? It is not FOL+SET!

[Mostowski Collapse]

Why is the Russell Paradox viewed as an application of the
Diagonal Argument? He he, Alzheimer wonky man, so you don't
remember? Your brains totally blank, eaten by Corona Virus?

Because one can prove the Russell Paradox from a particular
instance of the Subset axiom schema:

∀k∃u(Su ∧ ∀a(Eau ↔ (Eak ∧ ¬Eaa))) → ∀s(Ss → ∃a¬Eas) is valid.
https://www.umsu.de/trees/#~6k~7u%28Su~1~6a%28Eau~4%28Eak~1~3Eaa%29%29%29~5~6s%28Ss~5~7a~3Eas%29

And this instance with ~Eaa corresponds to the construction
of an Anti-diagonal, as used in the Diagonal Argument. The
Regularity Axiom doesn't come into play, it is even not built-in

in DC Proof. See also here, similar Diagonal Argument used in Cantors Proof:

https://en.wikipedia.org/wiki/Cantor's_diagonal_argument#Uncountable_set

The Drinker Paradox doesn't need this argument. It
even doesn't need SET, and is also not supposed to use SET.
You can much more easily prove the Drinker Paradox

in FOL, by proof by cases:

"The proof begins by recognising it is true that either
everyone in the pub is drinking (in this particular round of drinks),
or at least one person in the pub isn't drinking."
https://paradox.fandom.com/wiki/Drinker_paradox

[Mostowski Collapse]

But the original Russell Paradox was not like this, it
wasn't using the non-naive set-builder. It used the naive
set-builder, and was communicated as such to Frege,

showing the Frege system inconsistent. You can also
use the naive set-builder to derive the outside element
theorem. But calling the non-naive version Drinker

Paradox, is of course a wonky man fallacy:

∃u∀a(Eau ↔ ¬Eaa) → ∀s∃a¬Eas is valid.
https://www.umsu.de/trees/#~7u~6a%28Eau~4~3Eaa%29~5~6s~7a~3Eas

But more shockingly for Frege, you can derive:

∀a(Eau ↔ ¬Eaa) → (Euu ∧ ¬Euu) is valid.
https://www.umsu.de/trees/#~6a%28Eau~4~3Eaa%29~5Euu~1~3Euu

See also here:

"Russell discovered the paradox in May or June 1901.
By his own account in his 1919 Introduction to Mathematical
Philosophy, he "attempted to discover some flaw in Cantor's

proof that there is no greatest cardinal. In a 1902 letter, he
announced the discovery to Gottlob Frege of the paradox
in Frege's 1879 Begriffsschrift."
https://en.wikipedia.org/wiki/Russell's_paradox#History

But he communicated a version with functions, how is this done?

[Mostowski Collapse]

Ha Ha, thats a wooping 23 years:

In a 1902 letter, he announced the discovery to Gottlob Frege
of the paradox in Frege's 1879 Begriffsschrift."
https://en.wikipedia.org/wiki/Russell's_paradox#History

Still more than the 18 years that wonky man is wasting
with DC Proof! Problems problems like with function spaces,
and from when is his Drinker Paradox blog?

So I guess wonky man will not surrender, we will still hear
more nonsense from wonky man, for quite some time!

[Michel Marconi]

Mostowski Collapse wrote:

> Dan Christensen is the uber-crank. He not only beats John Gabriel, he is
> on par with "John Garbiel: look I reinvented calculus, here is new
> calculus", in that he has "Dan Christensen: look I reinvented
>
> logic, here is DC Proof". He also beats Archimedes Plutonium,
> in that he has more Alzheimer than Archimedes Plution. You can talk to
> Dan Christensen and he doesn't understand acronyms:

you guys are dindoo nothen nigga.

[Dan Christensen]

On Sunday, October 9, 2022 at 5:26:54 AM UTC-4, Mostowski Collapse wrote:

[snip childish abuse]

> "I reinvented logic, here is DC Proof".

I have simply formalized the rules of logic that it seems mathematicians actually use in proofs. Must be frustrating as hell for you, Jan Burse.

[Dan Christensen]

On Sunday, October 9, 2022 at 5:52:07 AM UTC-4, Mostowski Collapse wrote:

[snip]

> > Drinker Paradox is a FOL Paradox

As I have shown, it is also a "paradox" of set theory. Actually it is a special case of the theorem ALL(s):[Set(s) => EXIST(x):[x in s => Q]]

Deal with it, Jan Burse.

[Mostowski Collapse]

How should ALL(s):[Set(s) => EXIST(x):[x in s => Q] even relate
to the Drinker Paradox when Q is not ALL(y):D(y) and x in s is not D(x)?

If it were a Russel Paradox as well, it would be listed here:

-The barber with "shave".
-The Grelling–Nelson paradox with "describer":
-Richard's paradox with "denote":
- "I am lying.", namely the liar paradox and Epimenides paradox,
- Russell–Myhill paradox
-The Burali-Forti paradox,
-The Kleene–Rosser paradox,
-Curry's paradox
-The smallest uninteresting integer paradox
-Girard's paradox
https://en.wikipedia.org/wiki/Russell's_paradox#Related_paradoxes

I don't see any Drinker Paradox in this list.

LMAO!

[Ross A. Finlayson]

All one "Liar".

[Mostowski Collapse]

Why don't you ask here:

Mathematics Stack Exchange
https://math.stackexchange.com/

MathOverflow
https://mathoverflow.net/

Whether there is anybody on this planet that believes
that your formula ALL(s):[Set(s) => EXIST(x):[x in s => Q]

is the Drinker Paradox. Could be fun!

[Michel Marconi]

Mostowski Collapse wrote:

> Why don't you ask here:
> Mathematics Stack Exchange https://math.stackexchange.com/
> MathOverflow https://mathoverflow.net/
> Whether there is anybody on this planet that believes that your formula
> ALL(s):[Set(s) => EXIST(x):[x in s => Q]

nonsense. And irrelevant. You are a khakhole khazar nazi. A goy. You try to
hide behind hitler, but you are his nazis, fighting the Russian people
1944. You disgusting pig.

Ukrainian army’s commander-in-chief poses in swastika bracelet
https://www.rt.com/russia/564358-ukraine-commander-swastika-bracelet/

[Mostowski Collapse]

Ha Ha, micro penis is a slow thinker. We told him
already that www.rt.com is blocked. So what does he post?

[Dan Christensen]

On Sunday, October 9, 2022 at 4:24:04 PM UTC-4, Mostowski Collapse wrote:
> How should ALL(s):[Set(s) => EXIST(x):[x in s => Q] even relate
> to the Drinker Paradox when Q is not ALL(y):D(y) and x in s is not D(x)?
>

Perhaps you didn't know, but it is often possible to formalize a problem in more than one way. Really! DP can be formalized with sets (as I have shown), predicates or a combination of both .

> If it were a Russel Paradox as well, it would be listed here:
>
> -The barber with "shave".
> -The Grelling–Nelson paradox with "describer":
> -Richard's paradox with "denote":
> - "I am lying.", namely the liar paradox and Epimenides paradox,
> - Russell–Myhill paradox
> -The Burali-Forti paradox,
> -The Kleene–Rosser paradox,
> -Curry's paradox
> -The smallest uninteresting integer paradox
> -Girard's paradox
> https://en.wikipedia.org/wiki/Russell's_paradox#Related_paradoxes
>
> I don't see any Drinker Paradox in this list.
>

AFAIK this purely set-theoretic resolution of DP is an original insight. Not a very profound insight, but original nevertheless.

Have you ever had an original insight, Jan Burse?

[Mostowski Collapse]

Perhaps you didn't know, but it is most often not possible to

formalize a problem in more than one way. Really!

For example set theory has no universal set, whereas
an universal predicate is not inconsistent:

ALL(x):U(x)

Doesn't lead to an inconsistency. It has a simple
model, namely U = V.

[Mostowski Collapse]

Finally, after a painful search, found a counter
example to Dan Christensens Drinker Paradox,

where he proofs that there is at least one non-drinker:

5 EXIST(a):~a e drinkers
Detach, 4, 2
http://www.dcproof.com/DrinkersThm1.htm

In this pub, I guess everybody is drinking:

Kerch Bridge on fire na na na na na
https://www.youtube.com/watch?v=xjy0vRvUsi8