Re: Anyone that disagrees with this is not telling the truth ---- V4

On 8/19/24 10:17 PM, olcott wrote:

On 8/19/2024 9:15 PM, Richard Damon wrote:

On 8/19/24 10:06 PM, olcott wrote:

On 8/19/2024 8:55 PM, Richard Damon wrote:

On 8/19/24 9:38 PM, olcott wrote:

On 8/19/2024 8:24 PM, Richard Damon wrote:

On 8/19/24 8:50 PM, olcott wrote:

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void DDD()
{
   HHH(DDD);
   return;
}
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_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]
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*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)

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No, it is just a basic fact that NO emulator can properly emulate the about input past the 4th instruction, as the data just isn't provided.
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The data is provided, but I did not say that the data
is not provided to an ordinary regular bath tub rubber
duck on a big blue sticker, and I won't.

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WHERE is it provided?
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You said only what was EXPLICITLY stated could be used.
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 Not any more I took that part out.

No you didn't, it is still in the base message of the thread.
To change the foundation of an arguement you need to make a CLEAR and obvious break from your previous on.
But then, you never understood that ever, because that requires an understanding of how truth works.
You don't understand that you can't just change things in the middle.
Also, it can't be DDD without the n and have different HHHn with the n as parts of them.
So, your claim is still ligically invalid.
Sorry, you just don't understand what you are talking about and proving you are nothing but a stupid liar≥

 

Sorry, you are just proving that you are jus a stupid liar.
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PERIOD.
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Note, above you say HHH, below you say HHHn, thus they are not related problems,
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Sorry, you are just proving yourself to be an ignorant liar.
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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
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The above claim boils down to this: (X ∧ Y) ↔ Z
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void EEE()
{
   HERE: goto HERE;
}
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HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.
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Nope, and HHHn creates a DDDn that will call HHHn that will return to it after emulating n steps, and then DDDn will halt
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You are just proving you don't understand the difference between the behavior of the actuall DDDn to the partial simulation of it by HHHn.
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DDDn's behavior doesn't stop just because the emulator looking at it stopped, but continues to the final state,
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The Emulation of DDDn by HHHn DOES stop when HHHn stops emulating, and thus doesn't tell you what happens afterwards unless you can form an actual valid inductive argument to carry the emulation forward.
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Sorry, you are just proving how stupid you are.
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And, that you are nothing but a cheat that trys to twist the words.
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