Re: Olcott is simply wrong --- Try to prove otherwise --- pinned down

On 6/2/24 10:19 AM, olcott wrote:

On 6/2/2024 6:51 AM, Richard Damon wrote:

On 6/1/24 11:33 PM, olcott wrote:

On 6/1/2024 6:27 PM, Richard Damon wrote:

On 6/1/24 7:12 PM, olcott wrote:

On 6/1/2024 6:02 PM, Richard Damon wrote:

On 6/1/24 6:40 PM, olcott wrote:

>
Show me where I said anything in the above spec about an aborted simulation.

>
So, why did HH stop simulating after some n steps?
>
Did it reach a final state in the simulation? if not, it ABORTED its simulation.
>

>
When every possible which way DD correctly simulated by HH never reaches
past its own simulated line 03 then

>
And a simulation either goes until it reaches a final state of the machine it is simulating, or it aborted its simulation.
>

>
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int HH(ptr p, ptr i);
01       int DD(ptr p)
02       {
03         int Halt_Status = HH(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
>
When every DD correctly simulated by any HH cannot possibly reach
past its own simulated line 03 in 1 to ∞ steps of correct simulation
of DD by HH then we have exhaustively examined every possible HH/DD
pair and each element has of this infinite set has the same property.

>
So?
>
It doesn't matter how many aborted simulaiton you do of a given input (and each HH simulated a DIFFERENT input since it simulated the INSTANCE of the template with a different HH)
>

>
In other words one cannot prove that every five pound rock weighs
more than every three pound rock, one must weigh them one-at-a-time?

>
Nope. But you need to show that each rock IS a five pound rock.
>
IF you weigh one rock, and find it is 5 pounds, doesn't mean that anothoer rock  rock that looks about the same is also 5 pouds,
>
You do seem to like you Herring in Red sauce, don't you.
>
The comparison here is that you have only "weighed" a very few of your DDs, only those built on an HH that NEVER aborts have been determined to not halt. The others are just haven't-yet-halted-after-n-steps, but we actually DO know that they WILL Halt after more.
>

>

The ONLY simulation that actually showed that ITS input was no-halting was the HH that never aborted, and it didn't answer.
>
Every other HH has a DIFFERENT INPUT and would be LYING to say it had that other input.
>

>
In other words (because each rock is different) one cannot prove that every five pound rock weighs more than every three pound rock, one must weigh them one-at-a-time?

Nope, unless of course you still need to weight them to show they ARE 5 pound rocks.
>

>
Every HH/DD pair of the infinite of every possible HH/DD pair
DD correctly simulated by HH NEVER HALTS.

>
That isn't even your original claim you were asking about.
>
Your claim wasn't about "Halting" because that is easily disproven, but that there correct PARTIAL simulation done by H never reaches the statement after the call.
>
You are just showing your true colors, that you just don't understand what you are talkinag about and get your lies confused.
>

>

>
*THIS PROVES THAT THE INPUT TO H(DD,DD) DOES NOT HALT*
*THIS PROVES THAT THE INPUT TO H(DD,DD) DOES NOT HALT*
*THIS PROVES THAT THE INPUT TO H(DD,DD) DOES NOT HALT*

>
Nope. Aborted simulation don't prove anything.
>

>
When for each element of the infinite set of every HH/DD pair DD
correctly simulated by HH cannot get past its own simulated line 03
then we know that none of the DD inputs to each HH(DD,DD) ever halts.

>
>
Nope. Try to actually PROVE that.
>

 Semantic tautologies are self-evident truth that prove themselves.
It is a fact that every five pound rock weights more than any
three pound rock. No need to weigh any rocks.

Right, so you don't need to weigh a five pound rock to know it is five bpounds.

 typedef int (*ptr)();  // ptr is pointer to int function in C
00       int HH(ptr p, ptr i);
01       int DD(ptr p)
02       {
03         int Halt_Status = HH(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
 Likewise we correctly deduce that for every HH/DD pair of the
infinite set of all HH/DD pairs that match the above template
every DD correctly simulated by HH never reaches past its own
simulated line 03, thus never halts.

Maybe the SIMULATION never reaches past it, but that doesn't mean the input (or the simulation) doesn't halt.
If H aborts its simulation, then the "simulation" reached its final state, that of being aborted. Yes, the machine simulated wasn't simulated to a final state, but the simulation did end.
And since H aborted its simulation, there is no evidence that the machine simulated will never reach a final state, and in fact, we can show that
So, your "Self-Evident Truth" is proved to be incorrect, just like so many of your "self-evident truths" because

 *WHEN WE LOOK AS THE X86 MACHINE CODE OF DD THIS IS UNEQUIVOCAL*
DD correctly emulated by HH with an x86 emulator cannot possibly
reach past its own machine instruction [00001c2e] in any finite (or
infinite) number of steps of correct emulation.

Wrong.
With my non-pure HH, we can easily construct an example which does correctly simulate the x86 instructions given to it to the end
And, with a pure HH, we can positively PROVE that while HH may not be able to do its own "correct simulation" to that point, if this HH is programed to ever abort this simulation and return 0, we CAN prove that an ACTUAL COMPLETE CORRECT SIMULATION of this input (by a different program that actually does the complete simulation) will reach the final state, and thus the behavior specified by the input is Halting.
Thus, NO HH can be "correct" to actually say its input is non-halting, as the act of doing so make itself wrong.

 _DD()
[00001c22] 55         push ebp
[00001c23] 8bec       mov ebp,esp
[00001c25] 51         push ecx
[00001c26] 8b4508     mov eax,[ebp+08]
[00001c29] 50         push eax        ; push DD 1c22
[00001c2a] 8b4d08     mov ecx,[ebp+08]
[00001c2d] 51         push ecx        ; push DD 1c22
[00001c2e] e80ff7ffff call 00001342   ; call HH
[00001c33] 83c408     add esp,+08
[00001c36] 8945fc     mov [ebp-04],eax
[00001c39] 837dfc00   cmp dword [ebp-04],+00
[00001c3d] 7402       jz 00001c41
[00001c3f] ebfe       jmp 00001c3f
[00001c41] 8b45fc     mov eax,[ebp-04]
[00001c44] 8be5       mov esp,ebp
[00001c46] 5d         pop ebp
[00001c47] c3         ret
Size in bytes:(0038) [00001c47]