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

On Fri, 2024-06-07 at 15:01 -0500, olcott wrote:

On 6/7/2024 2:43 PM, wij wrote:

On Fri, 2024-06-07 at 14:31 -0500, olcott wrote:

On 6/7/2024 1:57 PM, wij wrote:

On Fri, 2024-06-07 at 13:41 -0500, olcott wrote:

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 ]
 
In comp.theory olcott <polcott333@gmail.com> wrote:
 
[ .... ]
 

_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.
 
That now seems to have changed.
 

 
Because I am a relatively terrible writer I must constantly
improve my words on the basis of reviews.
 
Try to show how this DD correctly simulated by any HH ever
stops running without having its simulation aborted by HH.
 
_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.
 
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.
 

 
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).
 

 
The Halting Problem asks for a program H (precisely a TM) that:
IF H(D,D)==1, THEN D(D) will return.
ELSE If H(D,D)==0, THEN D(D) will never return.
ELSE HP is undecidable
 
You keep solving POOH !!! and made lots of lies.
 
Surrender to my GUR, son.
 

 
If people are going to be dishonest about simple things
such as the actual behavior of actual x86 code where
they consistently deny verified facts
 
then we certainly cannot trust these people with more
difficult issues that require at least some slight degree
of judgment call.
 
When we can show that even in the halting problem HH
is only required to report on the behavior of DD correctly
simulated by HH these dishonest people merely use that
as another deflection point for their dishonesty.
 
The way around this that just worked is to stay diligently
focused one one single point until the dishonest people
finally admit that they have simply ignored all the proofs
for three solid years.
 
On 6/5/2024 10:58 PM, Richard Damon wrote:
  > On 6/5/24 11:44 PM, olcott wrote:
  >>
  >> THIS IS ALL THAT YOU WILL EVER GET TO TALK
  >> TO ME ABOUT UNTIL YOU ACKNOWLEDGE THAT
  >> I AM CORRECT OR YOU PROVE THAT I AM INCORRECT
  >
  > But, as I said, I won't acknowledge that you
  > are correct, because I am not willing to put
  > that effort into your worthless claim.
  >
 
Here is the earliest version of the proof (that everyone
has simply ignored for three solid years) that P correctly
simulated by H would never stop running unless aborted.
 
Halting problem undecidability and infinitely nested simulation
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
 
 
 
The fact that the execution trace of P derived by the executed
H and the simulated H exactly matches the machine code of P
proves that each instruction of P was simulated correctly and
in the correct order this conclusively proves that P is correctly
simulated by both of these instances of H.
 
It has proved this since 2021-09-26 and everyone has made
sure to ignore this proof so that they can maintain their false
assumption.
 
 
 
Try to show how this DD correctly simulated by any HH ever
stops running without having its simulation aborted by HH.
 
_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.
 
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.
 

 
I recalled now. You knew what the Halting Problem is. But soon, you started to insist
POOH is correct by fabricating ...(lots of things)...
Wake up, your trick won't work.
 

 
All the while that you insist on lying about
 
the verified fact that DD correctly simulated by
HH cannot possibly stop running without having its
simulation aborted by HH.
 
You won't get to talk to me about anything else such
as why the above statement matters.
 
If you want to choose to be a liar thus <is> the
consequence that you get.
 

The Halting Problem asks for a program H (precisely a TM) that:
IF H(D,D)==1, THEN D(D) will return.
ELSE If H(D,D)==0, THEN D(D) will never return.
ELSE HP is undecidable

What is your answer? (the last time I saw POOH, it does not report anything)