Re: DDD correctly emulated by H0 -- Ben agrees that Sipser approved criteria is met

On 6/26/24 8:20 PM, olcott wrote:

On 6/26/2024 6:55 PM, Richard Damon wrote:

On 6/26/24 7:46 PM, olcott wrote:

On 6/26/2024 6:41 PM, Richard Damon wrote:

On 6/26/24 9:42 AM, olcott wrote:

On 6/26/2024 6:02 AM, Richard Damon wrote:

On 6/25/24 11:42 PM, olcott wrote:

>
That is not the way that it actually works.
That the the way that lies are defined.

>
Source for you claim?
>
Where is you finite set of steps from the truthmakers of the system to that claim?
>

>
_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]
>
The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.

>
Sure it can. I have shown an H0 that does so.
>

>
I already told you that example does not count.
>
I can't keep repeating those details or others
that so far have no idea what an x86 emulator is
will be baffled beyond all hope of comprehension.
>

>
WHy not?
>

 We have already been over that you know that you cheated.
 

Nope, since you didn't put in the rule, and if you had it would have shown that you lied, as if H0 is a pure function then the call to H0 emulated by H0 needs to have the same behaivor as the direct call to H0 by main.
Since it doesn't, either H0 is incorrect, or you don't have that rule (and your claim about H0 is incorrect).