Re: Anyone that disagrees with this is not telling the truth --- V5

On 2024-08-20 13:13:57 +0000, olcott said:

On 8/20/2024 3:45 AM, joes wrote:

Am Mon, 19 Aug 2024 23:33:52 -0500 schrieb olcott:

On 8/19/2024 11:02 PM, Richard Damon wrote:

On 8/19/24 11:50 PM, olcott wrote:

On 8/19/2024 10:32 PM, Richard Damon wrote:

On 8/19/24 10:47 PM, olcott wrote:

*Everything that is not expressly stated below is*
*specified as unspecified*

Looks like you still have this same condition.
I thought you said you removed it.

 

_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]

But it can't emulate DDD correctly past 4 instructions, since the 5th
instruciton to emulate doesn't exist.
And, you can't include the memory that holds HHH, as you mention HHHn
below, so that changes, but DDD, so the input doesn't and thus is
CAN'T be part of the input.

Changing the code, but not the address, constitutes a change.
 

x86utm takes the compiled Halt7.obj file of this c program
https://github.com/plolcott/x86utm/blob/master/Halt7.c Thus making
all of the code of HHH directly available to DDD and itself. HHH
emulates itself emulating DDD.

 Which is irrelevent and a LIE as if HHHn is part of the input, that
input needs to be DDDn
And, in fact,
Since, you have just explicitly introduced that all of HHHn is
available to HHHn when it emulates its input, that DDD must actually
be DDDn as it changes.
 Thus, your ACTUAL claim needs to be more like:
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
The above claim boils down to this: (X ∧ Y) ↔ Z
 

Yes that is correct.

 So, you only prove that the DDD∞ that calls the HHH∞ is non-halting.
Not any of the other DDDn

 

Your problem is that for any other DDDn / HHHn, you don't have Y so
you don't have Z.

 

HHHn correctly predicts the behavior of DDD the same way that HHHn
correctly predicts the behavior of EEE.
 

Nope, HHHn can form a valid inductive proof of the input.
It can't for DDDn, since when we move to HHHn+1 we no longer have
DDDn but DDDn+1, which is a different input.
 

You already agreed that (X ∧ Y) ↔ Z is correct.
Did you do an infinite trace in your mind?

 But only for DDD∞, not any of the other ones.

 

If you can do it and I can do it then HHH can do this same sort of
thing. Computations are not inherently dumber than human minds.
 

But HHHn isn't given DDD∞ as its input, so that doesn't matter.
 

All of the DDD have identical bytes it is only the HHH that varies.
HHHn(DDD) predicts the behavior of HHH∞(DDD).
It does this same same way that HHHn(EEE)
predicts the behavior of HHH∞(EEE).

The bytes of HHH are part of DDD.
 

 *The following criteria only means*
HHHn(DDD) correctly predicts the behavior of HHH∞(EEE) and HHH∞(DDD)
 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then

No, it does not. You have not proven that H determines with a sound
method that its simulated D would never stop running unless aborted.
Nothe that above D is called H's input so it is important to apply
the correct meaning of the word "input".
--
Mikko