Re: At least 100 people kept denying the easily verified fact --- last communication with Richard

On 6/9/2024 1:08 PM, Richard Damon wrote:

On 6/9/24 10:04 AM, olcott wrote:

On 6/9/2024 4:33 AM, Mikko wrote:

On 2024-06-08 13:06:06 +0000, olcott said:
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On 6/8/2024 1:58 AM, Mikko wrote:

On 2024-06-07 18:41:47 +0000, olcott said:
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On 6/7/2024 1:24 PM, Richard Damon wrote:

On 6/7/24 2:02 PM, olcott wrote:

On 6/7/2024 12:50 PM, Alan Mackenzie wrote:

[ Followup-To: set ]
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In comp.theory olcott <polcott333@gmail.com> wrote:
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[ .... ]
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_DD()
[00001e12] 55         push ebp
[00001e13] 8bec       mov  ebp,esp
[00001e15] 51         push ecx
[00001e16] 8b4508     mov  eax,[ebp+08]
[00001e19] 50         push eax      ; push DD
[00001e1a] 8b4d08     mov  ecx,[ebp+08]
[00001e1d] 51         push ecx      ; push DD
[00001e1e] e85ff5ffff call 00001382 ; call HH

>

A {correct simulation} means that each instruction of the
above x86 machine language of DD is correctly simulated
by HH and simulated in the correct order.

>
That's a bit of sudden and substantial change, isn't it?  Less than a few
days ago, you were defining a correct simulation as "1 to N instructions"
simulated (without ever specifying what you meant by N).  It seems that
the simulation of exactly one instruction would have met your criterion.
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That now seems to have changed.
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Because I am a relatively terrible writer I must constantly
improve my words on the basis of reviews.
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Try to show how this DD correctly simulated by any HH ever
stops running without having its simulation aborted by HH.
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_DD()
[00001e12] 55         push ebp
[00001e13] 8bec       mov  ebp,esp
[00001e15] 51         push ecx
[00001e16] 8b4508     mov  eax,[ebp+08]
[00001e19] 50         push eax      ; push DD
[00001e1a] 8b4d08     mov  ecx,[ebp+08]
[00001e1d] 51         push ecx      ; push DD
[00001e1e] e85ff5ffff call 00001382 ; call HH
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A {correct simulation} means that each instruction of the
above x86 machine language of DD is correctly simulated
by HH and simulated in the correct order.
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Anyone claiming that HH should report on the behavior
of the directly executed DD(DD) is requiring a violation
of the above definition of correct simulation.
>

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And thus you admit that HH is not a Halt Decider,

>
More dishonest deflection.
The point that I made and you try to deflect using the strawman
deception as a fake rebuttal is the I just proved that DD is correctly
simulated by HH and this is not the same behavior as the directly
executed DD(DD).

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The true point is that you have never shown any proof about simulation
by HH.

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In other words you lack the mandatory prerequisites so the
correct proof only looks like gibberish to you.

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Hard to teest wihout the correct proof.
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Anyway, something that starts with "Proof:" and ends with "Q.E.D." may
fail to be a proof. It depends on what is between.
>

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typedef void (*ptr)(); // pointer to void function
>
void HHH(ptr P, ptr I)
{
   P(I);
   return;
}
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void DDD(int (*x)())
{
   HHH(x, x);
   return;
}
>
int main()
{
   HHH(DDD,DDD);
}
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In the above Neither DDD nor HHH ever reach their own return
statement thus never halt.
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When HHH is a simulating halt decider then HHH sees that
DDD correctly simulated by HHH cannot possibly reach its
own return statement, AKA

 But HHH (as shown above) ISN'T a simulating halt decider, so you are just caught in another of youre lies.
 

I didn't say it was a simulating halt decider. I needed to see
if my reviewers understand what infinite recursion is before
proceeding with them. It looks like they do not understand this.
That you call everything that I say a lie even when it
is not even incorrect is ridiculously childish of you.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer