Olcott can't answer this barber question

[immibis]

True or false?
∀x( ({{x}} ∈ Shaves) ⇔ ¬({{Barber},{Barber,x}} ∈ Shaves) )

Note: {{x},{x,y}} is the Kuratowski definition of an ordered pair in
ZFC, normally abbreviated as (x,y) however (x,y) is not defined syntax
in ZFC. If x=y this is the same as {{x}}.

Note: "Shaves" is a set of Kuratowski ordered pairs representing all
shaving relationships in the universe. If {{x},{x,y}} ∈ Shaves then it
means x shaves y. If ¬({{x},{x,y}} ∈ Shaves) then it means x does not
shave y.

[immibis]

I still don't see any answer from Olcott, despite many other messages
from him. I suppose he needs some time to consider why he is wrong, or
maybe he thinks that ignoring his wrongness makes it go away.

[olcott]

This is off topic.
I was only referring to the key precedent where ZFC
corrected the bogus definition of {set} in naive set theory
to eliminate the undecidability of Russell's paradox.

If ZFC can do that then the definition of the halting problem
can be corrected to make halting decidable.

Other than that I don't give a rat's ass about ZFC
or Russell's paradox.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

[immibis]

On 1/20/24 16:23, olcott wrote:
> On 1/20/2024 4:09 AM, immibis wrote:
>> On 1/19/24 21:48, immibis wrote:
>>> True or false?
>>> ∀x( ({{x}} ∈ Shaves) ⇔ ¬({{Barber},{Barber,x}} ∈ Shaves) )
>>>
>>> Note: {{x},{x,y}} is the Kuratowski definition of an ordered pair in
>>> ZFC, normally abbreviated as (x,y) however (x,y) is not defined
>>> syntax in ZFC. If x=y this is the same as {{x}}.
>>>
>>> Note: "Shaves" is a set of Kuratowski ordered pairs representing all
>>> shaving relationships in the universe. If {{x},{x,y}} ∈ Shaves then
>>> it means x shaves y. If ¬({{x},{x,y}} ∈ Shaves) then it means x does
>>> not shave y.
>>
>> I still don't see any answer from Olcott, despite many other messages
>> from him. I suppose he needs some time to consider why he is wrong, or
>> maybe he thinks that ignoring his wrongness makes it go away.
>
> This is off topic.
> I was only referring to the key precedent where ZFC
> corrected the bogus definition of {set} in naive set theory
> to eliminate the undecidability of Russell's paradox.

By doing so, you showed you have no idea what you're talking about.

>
> If ZFC can do that then the definition of the halting problem
> can be corrected to make halting decidable.
>
> Other than that I don't give a rat's ass about ZFC
> or Russell's paradox.
>

Maybe you should try to understand why me asking this question does not
violate ZFC. It might help you understand something about the halting
paradox.

[Richard Damon]

On 1/20/24 10:23 AM, olcott wrote:
> On 1/20/2024 4:09 AM, immibis wrote:
>> On 1/19/24 21:48, immibis wrote:
>>> True or false?
>>> ∀x( ({{x}} ∈ Shaves) ⇔ ¬({{Barber},{Barber,x}} ∈ Shaves) )
>>>
>>> Note: {{x},{x,y}} is the Kuratowski definition of an ordered pair in
>>> ZFC, normally abbreviated as (x,y) however (x,y) is not defined
>>> syntax in ZFC. If x=y this is the same as {{x}}.
>>>
>>> Note: "Shaves" is a set of Kuratowski ordered pairs representing all
>>> shaving relationships in the universe. If {{x},{x,y}} ∈ Shaves then
>>> it means x shaves y. If ¬({{x},{x,y}} ∈ Shaves) then it means x does
>>> not shave y.
>>
>> I still don't see any answer from Olcott, despite many other messages
>> from him. I suppose he needs some time to consider why he is wrong, or
>> maybe he thinks that ignoring his wrongness makes it go away.
>
> This is off topic.
> I was only referring to the key precedent where ZFC
> corrected the bogus definition of {set} in naive set theory
> to eliminate the undecidability of Russell's paradox.
>
> If ZFC can do that then the definition of the halting problem
> can be corrected to make halting decidable.
>
> Other than that I don't give a rat's ass about ZFC
> or Russell's paradox.
>

ZFC didn't "Redefine" Naive Set Theory.

It created a new ZFC set theory.

If you want to talk about PO-Computation Theory, go ahead, just make
sure you qualify all your usages to make it clear you are not talking
about the accepted Computation Theory,

Note, Naive Set theory had been proven "broken" before ZFC was worked
on, and it was a hot topic to figure out how to create a new version of
set theory.

If you can actually show a similar problem with Computation Theory,
maybe you could get more traction, but this seems to b something you
can't do. You show how it creates concepts that you don't like, largely
because you don't actually understand them, but you haven't demonstrated
a inconsistancy anywhere like that of the Russel Paradox that showed
Naive Set Theory to be untenable.

[Mikko]

On 2024-01-20 15:23:13 +0000, olcott said:

> On 1/20/2024 4:09 AM, immibis wrote:
>> On 1/19/24 21:48, immibis wrote:
>>> True or false?
>>> ∀x( ({{x}} ∈ Shaves) ⇔ ¬({{Barber},{Barber,x}} ∈ Shaves) )
>>>
>>> Note: {{x},{x,y}} is the Kuratowski definition of an ordered pair in
>>> ZFC, normally abbreviated as (x,y) however (x,y) is not defined syntax
>>> in ZFC. If x=y this is the same as {{x}}.
>>>
>>> Note: "Shaves" is a set of Kuratowski ordered pairs representing all
>>> shaving relationships in the universe. If {{x},{x,y}} ∈ Shaves then it
>>> means x shaves y. If ¬({{x},{x,y}} ∈ Shaves) then it means x does not
>>> shave y.
>>
>> I still don't see any answer from Olcott, despite many other messages
>> from him. I suppose he needs some time to consider why he is wrong, or
>> maybe he thinks that ignoring his wrongness makes it go away.
>
> This is off topic.
> I was only referring to the key precedent where ZFC
> corrected the bogus definition of {set} in naive set theory
> to eliminate the undecidability of Russell's paradox.
>
> If ZFC can do that then the definition of the halting problem
> can be corrected to make halting decidable.
>
> Other than that I don't give a rat's ass about ZFC
> or Russell's paradox.

Russell's paradox is not undecidable in the naive type theory.
Both R ∈ R and R ∉ R can be proven.

Mikko