Re: Anyone that claims this is not telling the truth

On 8/17/24 11:37 PM, olcott wrote:

On 8/17/2024 5:21 PM, Richard Damon wrote:

On 8/17/24 6:17 PM, olcott wrote:

On 8/17/2024 5:09 PM, Richard Damon wrote:

On 8/17/24 5:43 PM, olcott wrote:

On 8/17/2024 4:11 PM, Richard Damon wrote:

On 8/17/24 5:05 PM, olcott wrote:

On 8/17/2024 4:00 PM, Richard Damon wrote:

On 8/17/24 4:41 PM, olcott wrote:

On 8/17/2024 3:20 PM, Richard Damon wrote:

On 8/17/24 4:10 PM, olcott wrote:

On 8/17/2024 2:27 PM, Richard Damon wrote:
 > On 8/17/24 3:00 PM, olcott wrote:>> On 8/17/2024 1:50 PM, Richard Damon wrote:
 >>>>>> And thus ALL of memory is part of the input,

>
Any additional details have no effect what-so-ever on my claim.
>

>
Suure it does.
>
Since your argument tries to say that since DDD is the same to all of them, so its the behavior.
>
You are just admitting to being a LIAR.
>

>
*Calling me a liar admits that insults is all that you have*
*If I made a mistake then show that*

>
I did.
>

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FOR THREE YEARS YOU ALWAYS CHEAT
BY CHANGING MY WORDS AND REBUTTING THESE CHANGED WORDS
>
*Everything that is not expressly stated below is*
*specified as unspecified*

>
Ok.
>

>
void DDD()
{
   HHH(DDD);
}
>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)
>

>
No, DDD can NOT be emulated accoreding to the semantics of the x86 langauge, because the contents of the location 000015d2 is not provided to be emulated, and will need to be emulated after emulating the call instruction.
>

>
Everything that is logically entailed by the above specification
is included by reference. The assumption that DDD and HHH were
not in the same memory space has always been ridiculous.
>

>
Then I guess you accept that every different HHH generates a DIFFERENT Input, as that input, BY LOGHICAL NECESSITY includes all the code of HHH so it can be emulated, and thus you claims that "All the DDDs have the same bytes" is just a blantent lie.
>

>
This is my only claim
*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)
>
I am not claiming anything about any bytes.
>
>

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And, as I point out, that isn't true if HHH ever aborts its simulation.
>

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That is merely agreeing with what I said
>
X = DDD emulated by HHH according to the semantics of the x86 language
Y = HHH never aborts its emulation of DDD
Z = DDD never stops running
>
I said: (X ∧ Y) ↔ Z
You said ~Y which entails ~Z just like I said.
>
I had to rewrite that a bunch of times.
>

>
But, that also means that you have agreed that this only hold is HHH doesn't EVER abort its emulaiton,

>
In the same way that
X = when you are starving hungry
Y = never eat
Z = you will die
(X ∧ Y) ↔ Z
remains true yet does not hold in the case of ~X ∨ ~Y.
>
You never actually refuted (X ∧ Y) ↔ Z
You simply started with ~Y.
>

>
Strawman, and category error.
>

 Trying to refute (X ∧ Y) ↔ Z
with ~Y is stupid and does not work.
 

IU WASN'T refuting (X ∧ Y) ↔ Z
IO was pointing out that "HHH never aborts its emulation of DDD" isn't the HHH that you are thinking of, and thus we don't have your Y.
This is shown by the fact you brought up the Termination Analyzers, which CAN'T be your HHH that doesn't abort.
You are just tilting strawmen.
Note, refuting the conclusion Z from (X ∧ Y) ↔ Z CAN be done by showing that ~Y is a thing.
This is especially true since you have shown that you think PARTIAL simulation are CORRECT simulations, and that you like to switch the meaning of the words, and you are going to change the words of:
It is a basic fact that DDD emulated by HHH according to
the semantics of the x86 language cannot possibly stop
running unless aborted (out of memory error excluded)
from having the unless aborted for applying to the HHH that DDD calls, which you seem to be admtting here, to being JUST about the outer HHH, to claim that even if HHH does abort its emulation, and thus the DDD that calls that HHH will become halting, it is correct to call it non-halting, because if we change this HHH to non-halting, which also changes the code of the input since you include it as part of the input because it is logically necessary, but want to ignore it for the definiton of the input being the same..
Because you have PROVEN that you WILL attempt such a move, being VERY EXPLICIT in what is said is important.
If it breaks your proof because it closes the needed wiggle room, that shows your proof was ALWAYS incorrect.