Wonky man aka Dan Christensen sophistry are typical
crank farts. When refering to the Russell Paradox we of
course refer to everything the Russell Paradox needs,

none of it is needed for the Drinker Paradox. Here is
another proof, again in FOL:

To add a proof in Fitch style, natural deduction:
19 | Ex(Px -> VyPy) 18 ~~ Elimination
https://math.stackexchange.com/a/807721

Not a single line towards Russell Paradox needed,
not as a theorem, not as something derived from it,
not any axiom from SET needed, its not a FOL+SET

proof, its simply a FOL proof. Something wonky man
aka Dan Christensen and his ant brain cannot grasp.
How is this possible he asks, since his crank tool

cannot do it. On the other hand his crank tool DC Spoiled
should be able to prove the following, this is also valid in an
empty domain. So why can wonky man not prove it

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

without resorting to the Russell Paradox and related stuff?

LMAO!

Mostowski Collapse schrieb am Sonntag, 16. Oktober 2022 um 15:50:58 UTC+2:
> Dan Christensen halucinated:
> > Let's see the proof with each line justified by citing a single axiom or rule of inference.
> Which inference rule is invalid in FOL?
>
> Dan Christensen halucinated:
> > Maybe you didn't know, but Russell's Paradox (the non-existence of a universal set) is not an axiom of set theory.
> What does it tell you that it can be proved in FOL, and doesn't need FOL+SET?
> Dan Christensen schrieb am Samstag, 15. Oktober 2022 um 21:21:27 UTC+2:
> > On Saturday, October 15, 2022 at 1:45:23 PM UTC-4, Mostowski Collapse wrote:
> > > If one can prove this without Russell Paradox (using Exy for x ∈ y):
> > >
> > > ALL(drinkers):EXIST(x):[x in drinkers => ALL(y):[y in drinkers]]
> > > ∀d∃x(Exd → ∀yEyd) is valid.
> > > https://www.umsu.de/trees/#~6d~7x%28Exd~5~6yEyd%29
> > Let's see the proof with each line justified by citing a single axiom or rule of inference. Not today? Oh, well...
> > > Then this is also provable without Russell Paradox,
> > > which has two weakenings in the ALL quantifiers (using Sx for Set(x))
> > > ALL(drinkers):ALL(pub):[Set(drinkers) & Set(pub) => EXIST(x):[x in
> > > drinkers => ALL(y):[y in pub => y in drinkers]]]
> > >
> > > ∀d∀p((Sd ∧ Sp) → ∃x(Exd → ∀y(Eyp → Eyd))) is valid.
> > > https://www.umsu.de/trees/#~6d~6p%28Sd~1Sp~5~7x%28Exd~5~6y%28Eyp~5Eyd%29%29%29
> > >
> > > Why do you need the Russell Paradox?
> > Maybe you didn't know, but Russell's Paradox (the non-existence of a universal set) is not an axiom of set theory. It is THEOREM of set theory, the only axiom of set theory used in its proof being the Subset Axiom (Specification in ZFC). You can prove some version DP using only predicate logic, but it seems to require a mysterious, overarching, non-empty domain of discussion. In conventional mathematics, we make no such assumptions. A set-theoretic version of DP is provable, the only axiom of set theory required being the Subset Axiom. Deal with it, Jan Burse.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

Interesting other "dual" of the Drinker Paradox is this one,
again a pure FOL proof available, still no FOL+SET needed,
contrary to what Wonky Man aka Dan Christensen wants:

∃x(∃yQy → Qx) is valid.
https://www.umsu.de/trees/#~7x%28~7yQy~5Qx%29

Its even more intuitively true, without invoking the
Russell Paradox and related stuff. There are two cases:

case 1: ∃yQy, well modell theoretically we can take any
witness y0, which did Qy0 make true, and make the quantified
main sentence also true, ∃yQy → Qy0, which makes main
sentence also true, ∃x(∃yQy → Qx).

case 2: ~∃yQy, then ∃x(∃yQy → Qx) is vacuously true.

Credits: https://math.stackexchange.com/a/4225776

Mostowski Collapse schrieb am Sonntag, 16. Oktober 2022 um 15:5
9:31 UTC+2:
> Wonky man aka Dan Christensen sophistry are typical
> crank farts. When refering to the Russell Paradox we of
> course refer to everything the Russell Paradox needs,
>
> none of it is needed for the Drinker Paradox. Here is
> another proof, again in FOL:
>
> To add a proof in Fitch style, natural deduction:
> 19 | Ex(Px -> VyPy) 18 ~~ Elimination
> https://math.stackexchange.com/a/807721
>
> Not a single line towards Russell Paradox needed,
> not as a theorem, not as something derived from it,
> not any axiom from SET needed, its not a FOL+SET
>
> proof, its simply a FOL proof. Something wonky man
> aka Dan Christensen and his ant brain cannot grasp.
> How is this possible he asks, since his crank tool
>
> cannot do it. On the other hand his crank tool DC Spoiled
> should be able to prove the following, this is also valid in an
> empty domain. So why can wonky man not prove it
>
> ALL(drinkers):EXIST(x):[x e drinkers => ALL(a):[a e drinkers]]
>
> without resorting to the Russell Paradox and related stuff?
>
> LMAO!
> Mostowski Collapse schrieb am Sonntag, 16. Oktober 2022 um 15:50:58 UTC+2:
> > Dan Christensen halucinated:
> > > Let's see the proof with each line justified by citing a single axiom or rule of inference.
> > Which inference rule is invalid in FOL?
> >
> > Dan Christensen halucinated:
> > > Maybe you didn't know, but Russell's Paradox (the non-existence of a universal set) is not an axiom of set theory.
> > What does it tell you that it can be proved in FOL, and doesn't need FOL+SET?
> > Dan Christensen schrieb am Samstag, 15. Oktober 2022 um 21:21:27 UTC+2:
> > > On Saturday, October 15, 2022 at 1:45:23 PM UTC-4, Mostowski Collapse wrote:
> > > > If one can prove this without Russell Paradox (using Exy for x ∈ y):
> > > >
> > > > ALL(drinkers):EXIST(x):[x in drinkers => ALL(y):[y in drinkers]]
> > > > ∀d∃x(Exd → ∀yEyd) is valid.
> > > > https://www.umsu.de/trees/#~6d~7x%28Exd~5~6yEyd%29
> > > Let's see the proof with each line justified by citing a single axiom or rule of inference. Not today? Oh, well...
> > > > Then this is also provable without Russell Paradox,
> > > > which has two weakenings in the ALL quantifiers (using Sx for Set(x))
> > > > ALL(drinkers):ALL(pub):[Set(drinkers) & Set(pub) => EXIST(x):[x in
> > > > drinkers => ALL(y):[y in pub => y in drinkers]]]
> > > >
> > > > ∀d∀p((Sd ∧ Sp) → ∃x(Exd → ∀y(Eyp → Eyd))) is valid.
> > > > https://www.umsu.de/trees/#~6d~6p%28Sd~1Sp~5~7x%28Exd~5~6y%28Eyp~5Eyd%29%29%29
> > > >
> > > > Why do you need the Russell Paradox?
> > > Maybe you didn't know, but Russell's Paradox (the non-existence of a universal set) is not an axiom of set theory. It is THEOREM of set theory, the only axiom of set theory used in its proof being the Subset Axiom (Specification in ZFC). You can prove some version DP using only predicate logic, but it seems to require a mysterious, overarching, non-empty domain of discussion. In conventional mathematics, we make no such assumptions. A set-theoretic version of DP is provable, the only axiom of set theory required being the Subset Axiom. Deal with it, Jan Burse.
> > > Dan
> > >
> > > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > > Visit my Math Blog at http://www.dcproof.wordpress.com

Mostowski Collapse wrote:

> Dan Christensen halucinated:
>> Maybe you didn't know, but Russell's Paradox (the non-existence of a
>> universal set) is not an axiom of set theory.
>
> What does it tell you that it can be proved in FOL, and doesn't need
> FOL+SET?

the people revolting now in France etc, say that you two nazis of
nonexistent "uKraine" will hang in trees along the roads, due your
activities for destruction the lives of people in the fucking europe.

sri-lanka, I read, was a big power in the past. Now they can't even produce
their food. This is what you did with gearmony, etc and the cacapitalist
cacamericacan europe. They wanted kill you, openly, with "vaccines" for
fuck sake.

Dan Christensen wrote:

> On Saturday, October 15, 2022 at 1:45:23 PM UTC-4, Mostowski Collapse
> wrote:
>> If one can prove this without Russell Paradox (using Exy for x ∈ y):
> Let's see the proof with each line justified by citing a single axiom or
> rule of inference. Not today? Oh, well...

moreover, you two nazis of "uKraine", a creation of CCCP 1990, I read the
Russian Federation Army left Kiev after the putina had an agreement of
peace with the khazar gay actor cocaine zelenske. So they left, as part of
the agreement, where after the zelenske lying bitch, instructed by the
wanker boris johnson, dismissed the agreement, deciding to attack the good
people of Russia. The putina is an imbecile, I'm forced to conclude.

Mostowski Collapse wrote:

> Dan Christensen aka Wonky Man halucinated:
>> My set-theoretic version does not require it
>
> How do you prove it without an non-empty domain?
> Can you exactly show us how the drinker paradox

khazar Ghislaine Maxwell calls khazar Prince Andrew ‘dear friend’
https://www.rt.com/news/564756-maxwell-andrew-friend-interview/