Re: DDD correctly emulated by EEE --- Correct Emulation Defined --- CUT-AND-PASTE FAILED

On 3/24/25 10:05 AM, olcott wrote:

On 3/24/2025 6:23 AM, Richard Damon wrote:

On 3/23/25 9:27 PM, olcott wrote:

On 3/23/2025 6:55 PM, Richard Damon wrote:

On 3/23/25 6:56 PM, olcott wrote:

On 3/23/2025 4:46 PM, Richard Damon wrote:

On 3/23/25 1:38 PM, olcott wrote:

On 3/23/2025 6:08 AM, Richard Damon wrote:

On 3/22/25 11:57 PM, olcott wrote:

On 3/22/2025 9:53 PM, Richard Damon wrote:

On 3/22/25 2:00 PM, olcott wrote:

On 3/22/2025 12:34 PM, Richard Damon wrote:

On 3/22/25 10:52 AM, olcott wrote:

_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call EEE(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
>
When finite integer N instructions of the above x86
machine language DD are emulated by each x86 emulator
EEE[N] at machine address [000015c3] according to the
semantics of the x86 language no DD ever reaches its own
"ret" instruction at machine address [00002155] and
terminates normally.
>

>
Your can't emulate the above code for N > 4, as you get into undefine memory.
>

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I have already addressed this objection dozens of times.
>

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No you haven't. You have given several different LIES about it.
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As I have pointed out, if you don't include Halt7.c as part of the definition, then you can't do it as you are looking at undefined memory.
>

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Your lack of technical competence is showing.
(1) We are talking about a hypothetical infinite
set of pure x86 emulators that have no decider code.
>
(2) The memory space of x86 machine code is not
in the C source file, it is in the object file.
>

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Then your "input" isn't the C source files, but the memory, and ALL of it, and thus in your (1), each member of the set got a different input (as reference memory changed) and none of those apply to your case with HHH.
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You just continue to prove that you don't understand the meaning of the terms you are using, or you are intentionally hiding your fradulant change of meaning of those terms.
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Command line arguments:
x86utm Halt7.obj > Halt7out.txt
>
All of the x86 functions remain at their same fixed
offset from the beginning of Halt7.obj

>
So?
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You still need to make the decision, is Halt7.c / Halt7.obj part of the INPUT to the decider, and thus either you can't change the code in it, or you need to consider each version a different input, or
>

>
_III()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push III
[0000217a] e853f4ffff call 000015d2 ; call EEE(III)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
In other words an infinite set of pure x86 emulators
with each one stored at machine address 000015d2
that can be called from the above fixed finite string
of machine code IS UTTERLY BEYOND ANYTHING THAT YOU
CAN POSSIBLY IMAGINE.
>
I don't buy it. You are  neither that stupid nor
that ignorant.
>

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You can't have two different programs in one memory location at the same time.
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CUT-AND-PASTE FAILED
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I will dumb it down for you.
Try to come up with one x86 emulator EEE at machine
address 000015d2 that emulates III according to the
semantics of the x86 language and this emulated III
reaches its own machine address 00002183.
>

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No, STRAWMAN ERROR. You are just having a logic failure.
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No, you are showing yourself to be dumb.
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You can't redefine what "Correct Emulation" means without loosing the ability to use it to answer the problem, as we can only look as emulations instead of the original machine BECAUSE the are defined to be the same.
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The big problem with your example is that the fact that there doesn't exist an EEE that can correct emulate this input to it final state, is that all this proves is that this sort of emulator can never "prove" that this sort of input is halting.
>

 It only takes III calling EEE twice in sequence with
no conditional branch instructions between 00002172 and
0000217a to prove that III specifies not haling behavior.

Which NEVER happens!!!!
III calles EEE which will EMULATE the code, and thus there is NEVER another call to EEE actually executed again.
You confuse the emulation of the code with the execution of the code, just like you confuse your fantasies of the world with reality.
You are just proving how utterly STUPID and IGNORANT you are.

 

We also have the fact that none of the inputs you are looking at actually match the inputs given to your deciders, because, as been shown, the code at 000015d2 must be included in the input, or your decider fails to be the "pure function" you agree it must be.
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And if it is, it won't match the code of the decider, since EEE is not the same program as HHH.
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So, all you have done is proven that your logic is based on lies, and on creating strawmen with those lies.
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Your credibility is now in that lake of fire that you will be joining in the not to distant future. Sorry, but that is the facts unless you make a radical change in your behavior, and admit your error.
>
But then, it seems you are just too ignorant to be able to see your error, because you brainwashed yourself into being ignorant out of a fear of being brainwashed by the truth.