XPost: comp.theory, sci.logic

On 8/13/25 9:40 PM, olcott wrote:

How many tests that are black in color are entirely
white in color and the answer must be a positive
integer and must come with proof that it is correct.

What time is it (yes or no) ?

Is this sentence true or false: "This sentence is not true" ?
The above is the basis for the Tarski undefinability theorem.

What correct value can a halt decider return on an input
that does the opposite of whatever the halt decider decides?

The correct answer is that no such *INPUT* can possibly exist.

But the halting problem isn't such a problem.

First, the behavior of a Turing Machine / Input Compbinate always has a
defined behavior, and thus DOES either halt or not.

Your thoughts here are just based on a category error of assuming that
the decider gets to "make up its mind" after the input is created.

The halting question IS that quesiton, what does the program given as
the input do, and at the point we ask it, the decider has already been
fixed in design, and thus its asnwer for this input is fixed, and thus
asking what answer COULD it give to be correct, is just wrong, it can
only give one answer, and that will be incorrect, but there is a correct answer.

Your problem is that the question YOU are trying to ask is the invalid
one, as it presumes a choice to be made that isn't there.

So, you are just showing you don't understand what you are talking about.

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XPost: comp.theory, sci.logic

On 14/08/2025 03:29, olcott wrote:

<snip>

When the halting problem shows that there is an
input that does the opposite of whatever the halt
decider decides then "IF" this *INPUT* actually exists
it would prove by contradiction that no universal
halt decider exists.

When no such *INPUT* actually exists then the
whole proof totally falls completely apart.

Nonsense. If no such input exists, it is a fault easily remedied.
All you have to do is write it. Then it exists.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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XPost: comp.theory, sci.logic

On 8/13/25 11:06 PM, olcott wrote:

On 8/13/2025 9:56 PM, dbush wrote:

On 8/13/2025 10:29 PM, olcott wrote:

On 8/13/2025 9:02 PM, dbush wrote:

On 8/13/2025 9:56 PM, olcott wrote:

On 8/13/2025 8:45 PM, dbush wrote:

On 8/13/2025 9:40 PM, olcott wrote:

How many tests that are black in color are entirely
white in color and the answer must be a positive
integer and must come with proof that it is correct.

Error: Assumes that something can be entirely black and entirely
white

What time is it (yes or no) ?

Error: Assumes that the answer can be yes or no

Is this sentence true or false: "This sentence is not true" ?
The above is the basis for the Tarski undefinability theorem.

Error: Assume that sentence can have a truth value

Yes and by saying that you have proven that you
understand the Liar Paradox much better than every
expert on the philosophy of logic in the world.
The very best expert in the sub field of truthmaker
maximalism said that the Liar Paradox might not
have a truth value.

They all understand that.

What you don't understand is that if you assume that a truth
predicate exists, then by performing a set series of truth
preserving operations we reach the conclusion that the liar paradox
does have a truth value.

I have the actual Tarski proof and it does not go
that way at all.

https://liarparadox.org/Tarski_275_276.pdf

That's exactly how it goes.  You just don't understand it, just like
you don't understand the halting problem proof.

If you totally ignore all the theory / metatheory stuff
it may superficially seem that way.

No, it is YOU that is ignoring things, like the definition of
non-halting, or what a Program is.

Therefore no truth predicate exists.

Once again, you're proving you don't understand proof by contradiction. >>>>

When the halting problem shows that there is an
input that does the opposite of whatever the halt
decider decides

So you start with the assumption that a halt decider exists, i.e. you
have an H that meets these requirements:


Given any algorithm (i.e. a fixed immutable sequence of instructions)
X described as <X> with input Y:

A solution to the halting problem is an algorithm H that computes the
following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly

then "IF" this *INPUT* actually exists

Yet it quit talking about the behavior of the input.
That sneaky bait-and-switch has fooled everyone
for 89 years.

But the "behavior of the input" is DEFINED to be the behavior of the
program it represents.

That is what "represents" means, something you don't seem to understand.

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