Re: The philosophy of computation reformulates existing ideas on a new basis ---repeat until understood

On 10/30/24 11:59 PM, olcott wrote:

On 10/30/2024 10:18 PM, Richard Damon wrote:

On 10/30/24 10:28 PM, olcott wrote:

On 10/30/2024 6:35 PM, Richard Damon wrote:

On 10/30/24 8:28 AM, olcott wrote:

On 10/30/2024 6:19 AM, Richard Damon wrote:

On 10/29/24 9:41 PM, olcott wrote:

On 10/29/2024 8:17 PM, Richard Damon wrote:

On 10/29/24 10:41 AM, olcott wrote:

On 10/29/2024 5:50 AM, Richard Damon wrote:

On 10/28/24 11:08 PM, olcott wrote:

On 10/28/2024 9:56 PM, Richard Damon wrote:

On 10/28/24 9:09 PM, olcott wrote:

On 10/28/2024 6:56 PM, Richard Damon wrote:

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It is IMPOSSIBLE to emulate DDD per the x86 semantics without the code for HHH, so it needs to be part of the input.
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*You seemed to be a totally Jackass here*
You are not that stupid
You are not that ignorant
and this is not your ADD
>
_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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At machine address 0000217a HHH emulates itself emulating
DDD without knowing that it is emulating itself.
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Then how did it convert the call HHH into an emulation of DDD again?
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When HHH (unknowingly) emulates itself emulating DDD this
emulated HHH is going to freaking emulate DDD.
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Did you think it was going to play poker?
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Which is what it would do, get stuck and fail to be a decider. It might figure out that it is emulating an emulating decider, at which point it knows that the decider might choose to abort its conditional emulation to return, so it needs to emulate further.
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Only by recognizing itself, does it have grounds to say that if I don't abort, it never will, and thus I am stuck, so I need to abort.
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Counter-factual. This algorithm has no ability to its own code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c // page 801
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*That people fail to agree with this and also fail to*
*correctly point out any error seems to indicate dishonestly*
*or lack of technical competence*
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DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
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No, it knows its own code because it rule for "No conditional branches" excludes that code.
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Are you really so stupid that you think this will help
DDD reach its own return instruction?

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DDD doesn't need any help to reach its own return instruction, as the HHH that it calls DOES abort and return to it.
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Are you really so stupid that you think you can keep getting
away with the strawman deception by changing the subject away
from DDD emulated by HHH?
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What strawman?
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I am just going to the defintions of the problem you claim to be solving.
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Do I have to repeat this a few hundred times in every post
so that you can remember from one post to the next?
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*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*
*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*
*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*

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But, you haven't removed yourself from the topic, so the definitions still apply.
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HHH is each element of the set of x86 emulators that emulates zero
to infinity steps of DDD including zero to infinity emulations of
itself emulating DDD.

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No, it isn't, because your published HHH is not a "set of programs".
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It is *A* progrtam.
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DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
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Which means HHH can't abort its emulaiton, or it fails to meet its requrements.
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And thus the ONLY HHH that meets yor requrements is a different one than presented, and that one returns to NOBODY.
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You can't try to redefine the terms until you clearly and public announce that you are leaving Computation Theory behind, and nothing you talk about can be broght back in.
>

 EVERYTHING NOT EXPRESSLY STATED IS EXPRESSLY EXCLUDED
UNLESS ENTAILED BY THE SEMANTICS OF THE X86 LANGUAGE
 void DDD()
{
   HHH(DDD);
   return;
}
 DDD emulated by any HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
 

So, you are just admitting that you don't know what you are talking about, since NOTHING after your disclaimer that we can only evalute everything by the "semantics of the x86 language" is in the x86 languge, and thus not meaningful.
Sorry, it is clear you don't understand what the words you are using mean, because it seems there is a big funny-mental filter between your brain and the world.
Even if you replace the C function DDD with its x86 code equivalenet, you can't evaluate your later claim, as the "call HHH" instruction in the x86 code for DDD means EXACTLY, and nothing else, to continue in the function HHH, which hasn't been given so can't be done.
And when you DO include the code for HHH, you don't get the meaning you want, because it STILL means do the code of HHH, ahd that doesn't let you say that "means" emulated DDD per the x86 langugage, but it only means that doing the steps of HHH, which by logic OUTSIDE that of the x86 language can be seen to be an emulation of DDD (but that emulation is never directly seen in the emulation of DDD).
Also, HHH is not then allowed to ever "abort" its emulation, and the semantics of the x86 language don't allow for that, as EVERY instruction includes as part of its semantics what the next instruction to process WILL be, so any stopping of the emulation is a violation of your claimed semantics.
If the code of the HHH that DDD calls happens to be code that aborts, then the correct x86 emulation of DDD, which now isn't being done by that HHH, but still REQUIRED by your claim, will show that the emulation WILL reach the end, so the only valid correction of your statement that means anything close to it is shown to be false.
Sorry, but you are just proving how stupid and ignorant you are.
You don't understand the difference between a program, which has behavior, and the partial emulation of said program, in part because you have shown to not understand what it "Truth" and what is "Knowledge", confusing the two and thus getting EVERYTHING wrong.