Re: Everyone on this forum besides Keith has been a damned liar about this point

On Mon, 09 Jun 2025 11:54:18 -0400, dbush wrote:
 

On 6/9/2025 11:49 AM, olcott wrote:

On 6/9/2025 10:34 AM, dbush wrote:

On 6/9/2025 11:29 AM, olcott wrote:

On 6/9/2025 10:06 AM, dbush wrote:

On 6/9/2025 10:55 AM, olcott wrote:

On 6/9/2025 6:55 AM, dbush wrote:

On 6/9/2025 12:15 AM, olcott wrote:

On 6/8/2025 10:42 PM, dbush wrote:

On 6/8/2025 11:39 PM, olcott wrote:

On 6/8/2025 10:32 PM, dbush wrote:

On 6/8/2025 11:16 PM, olcott wrote:

On 6/8/2025 10:08 PM, dbush wrote:

On 6/8/2025 10:50 PM, olcott wrote:

void DDD()
{
    HHH(DDD);
    return;
}
>
The *input* to simulating termination analyzer HHH(DDD)

>
No it's not, as halt deciders / termination analyzers work
with algorithms,

>
That is stupidly counter-factual.
>
>

That you think that shows that

>
My understanding is deeper than yours.
No decider ever takes any algorithm as its input.

>
But they take a description/specification of an algorithm,

>
There you go.
>

which is what is meant in this context.

>
It turns out that this detail makes a big difference.
>

And because your HHH does not work with the description/
specification of an algorithm, by your own admission, you're not
working on the halting problem.
>
>

HHH(DDD) takes a finite string of x86 instructions

>
>
Which you stated only includes the instructions of the function
DDD on multiple occasions (see below),

>
It is proven that you are a liar by the part of my reply that you
erased.
>
HHH(DDD) takes a finite string of x86 instructions that specify
that HHH simulates itself simulating DDD.
>
>

Then you admit that that finite string includes the machine code of
the function DDD, the machine code of the function HHH, and the
machine code of everything that HHH calls down to the OS level, and
that address 000015c3 is part of DDD?

>
I admit that:
(a) DDD correctly simulated by HHH,
(b) the directly executed DDD() and (c) the directly executed HHH()
WOULD NEVER STOP RUNNING UNLESS HHH ABORTS ITS SIMULATION OF DDD.
>
Because this is true it derives conclusive proof that the input to
HHH(DDD) specifies a non-halting sequence of configurations.
>
That people here disagree with self-evident truth seems to indicate
that people here are liars.
>
In epistemology (theory of knowledge), a self-evident proposition is
a proposition that is known to be true by understanding its meaning
without proof... https://en.wikipedia.org/wiki/Self-evidence
>
>
>

>
In other words, a non-answer.  I'll take that as a no.
>
And since your HHH doesn't work with algorithms (or their description
/ specification) as you've admitted, you're not working on the halting
problem.
>
>

You are far too sloppy in your interpretation of the meaning of words.
Also when I do provide an answer you simply ignore it.

>
>
Replying with something other than "yes" or "no" to a yes or no question
is not an answer.
>
>

The input to HHH(DDD)

Which only consists of the the instructions of the function DDD, as you
have admitted, you're not working with algorithms (or their description
/ specification) and therefore not working on the halting problem.
>
If you would just quit lying about that people might take you seriously.
>
>
>
On 5/13/2025 9:54 PM, dbush wrote:
  > On 5/13/2025 9:48 PM, olcott wrote:
  >> On 5/13/2025 8:31 PM, dbush wrote:
  >>> On 5/13/2025 9:27 PM, olcott wrote:
  >>>> On 5/13/2025 8:07 PM, dbush wrote:
  >>>>> On 5/13/2025 5:30 PM, olcott wrote:
  >>>>>> On 5/13/2025 6:43 AM, Richard Damon wrote:
  >>>>>>> On 5/13/25 12:52 AM, olcott wrote:
  >>>>>>>> *simulated D would never stop running unless aborted*
  >>>>>>>> or they themselves could become non-terminating.
  >>>>>>>
  >>>>>>> But you aren't simulating the same PROGRAM D that the original
  >>>>>>> was given.
  >>>>>>>
  >>>>>>>
  >>>>>> It is not supposed to be the same program.
  >>>>>
  >>>>> So you *explicitly* admit to changing the input.
  >>>>>
  >>>>>
  >>>> The finite string of DD is specific sequence bytes.
  >>>
  >>> Which includes the specific sequence of bytes that is the finite
  >>> string HHH
  >>>
  >>>
  >> No it does not. A function calls is not macro inclusion.
  >>
  >>
  > Then you admit that your HHH not deciding about algorithms and
  > therefore has nothing to do with the halting problem.
  >
  >
On 6/7/2025 10:56 AM, dbush wrote:
  > On 6/7/2025 10:54 AM, olcott wrote:
  >> On 6/7/2025 9:51 AM, dbush wrote:
  >>> On 6/7/2025 10:32 AM, olcott wrote:
  >>>> The next instruction of DDD that HHH emulates is at the machine
  >>>> address of 00002183.
  >>>>
  >>>> The next instruction of DDD that HHH1 emulates is at the machine
  >>>> address of 00002190.
  >>>
  >>> False.
  >>>
  >>> The next instruction of DDD that both HHH and HHH1 emulates is at
  >>> the machine address of 000015c3,
  >>
  >> *That is not an instruction of DDD*
  >> *That is not an instruction of DDD*
  >> *That is not an instruction of DDD*
  >> *That is not an instruction of DDD*
  >
  > In other words, you're not operating on algorithms.  And since the
  > halting problem is about algorithms, what you're working on has
  > nothing to do with the halting problem.
  >
  > If you would just be honest about the fact that you're not working on
  > the halting problem, people would stop bothering you.

 Machine code is a REPRESENTATION of an algorithm.
 /Flibble