Re: DDD correctly emulated by H0 --- Why Lie? -- Repeat until Closure

On 6/25/24 11:37 PM, olcott wrote:

On 6/25/2024 10:32 PM, Richard Damon wrote:

On 6/25/24 11:25 PM, olcott wrote:

On 6/25/2024 9:54 PM, Richard Damon wrote:

On 6/25/24 10:21 PM, olcott wrote:

On 6/25/2024 9:17 PM, Richard Damon wrote:

On 6/25/24 9:59 PM, olcott wrote:

On 6/25/2024 8:47 PM, Richard Damon wrote:

On 6/25/24 9:02 AM, olcott wrote:

On 6/25/2024 6:04 AM, Richard Damon wrote:

On 6/24/24 11:13 PM, olcott wrote:

On 6/24/2024 9:48 PM, Richard Damon wrote:

On 6/24/24 10:38 PM, olcott wrote:

On 6/24/2024 9:26 PM, Richard Damon wrote:

On 6/24/24 10:21 PM, olcott wrote:

On 6/24/2024 9:02 PM, Richard Damon wrote:

On 6/24/24 9:55 PM, olcott wrote:

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*We can get to that as soon as you reverse your lie*
*We can get to that as soon as you reverse your lie*
*We can get to that as soon as you reverse your lie*

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You still haven't shown where I lied, on where you don't like what I say.
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You said that D correctly simulated by H must
have the behavior of the directly executed D(D).

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Right, the steps that H sees are IDENTIAL to the steps of the directly executed D(D) until H stops its simulation,
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NOT ONE DIFFERENCE.
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Honest mistake or liar?
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The directly executed D(D) has identical behavior to
D correctly simulated by H1
*the call from D to H(D,D) returns*
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This is not the same behavior as
D correctly simulated by H
*the call from D to H(D,D) DOES NOT return*
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And what instruction did H's simulation differ from the direct executions trace?
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D correctly simulated by H
*the call from D to H(D,D) DOES NOT return*

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Which isn't "Behavior of the input"
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The "not happening" of something that could have happened except that the processing was stoped is NOT behavior.
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D correctly simulated by H1 --- Identical to D(D)
*the call from D to H(D,D) returns*
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Right, and it contains ALL of the behavior of the correct simulation of D by H, plus more.
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H doesn't see DIFFERENT behavior, just LESS, and that differnce isn't due to the input, but due to H.

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*These are not the same behaviors*
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(Assuming unlimited memory)
When 1 to a googolplex of steps of D are correctly simulated by H
*the call from D to H(D,D) NEVER RETURNS*

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Correction, 1 to a googleplex of steps if DIFFERENT Ds, each paired with a DIFFERENT H, when simulated by that H, the DIFFFERENT routines called by those DIFFERENT Ds to that DIFFERENT H(D,D) is never simulated to an end.
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_P()
[000020e2] 55               push ebp         ; housekeeping
[000020e3] 8bec             mov ebp,esp      ; housekeeping
[000020e5] 51               push ecx         ; housekeeping
[000020e6] 8b4508           mov eax,[ebp+08] ; parameter
[000020e9] 50               push eax         ; push parameter
[000020ea] 8b4d08           mov ecx,[ebp+08] ; parameter
[000020ed] 51               push ecx         ; push parameter
[000020ee] e82ff3ffff       call 00001422    ; call H(P,P)
[000020f3] 83c408           add esp,+08
[000020f6] 8945fc           mov [ebp-04],eax
[000020f9] 837dfc00         cmp dword [ebp-04],+00
[000020fd] 7402             jz 00002101
[000020ff] ebfe             jmp 000020ff
[00002101] 8b45fc           mov eax,[ebp-04]
[00002104] 8be5             mov esp,ebp
[00002106] 5d               pop ebp
[00002107] c3               ret
Size in bytes:(0038) [00002107]
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The call from D to H(D,D) cannot possibly return when D
is correctly simulated by any H that can possibly exist.
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Unless you say yes you are correct we cannot move on to
the next point.
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No, the call most definitinely DOES return, but that return is after the simulation ended.
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Maybe the real problem is that you have insufficient technical competence.

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Nope, that isn't the problem. I KNOW what I am talking about, as opposed to you who can't even write a simple Turing Machine.
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Your problem is that, strictly, by your definition of "Correct Simulation",

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The semantics of the x86 language objectively proves that I am correct.
Have you been faking your technical competence?

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Nope.
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Can you do better with this simpler example?
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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 H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
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The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.
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But the call will, just not in the simulation that your H0 does.
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OK so we are back to you being a freaking liar trying to get
away with contradicting the semantics of the x86 language.
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How does that contradictthe semantics of the x86 languge?
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If H0 is a decider, it will ALWAYS return an answer in finite time

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H0 is not even a decider yet. When you leap ahead you
diverge from the point at hand.
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THen it is allowed to be the impure function , so you r claim is proven false.
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 We have not even gotten that far yet.
We ARE ONLY TALKING ABOUT the behavior of DDD correctly
simulated by H0. AKA software engineering and NOT computer
science.
 

So, you have no grounds to exclude my version of H0, so your claim is just proven false.
int H0(ptr x) {
   static int flag = 0;
   if (flag) return 0;
   flag = 1;
...