Re: DDD emulated by HHH diverges from DDD emulated by HHH1

On 6/4/25 10:06 PM, olcott wrote:

On 6/4/2025 8:48 PM, dbush wrote:

On 6/4/2025 9:39 PM, olcott wrote:

On 6/4/2025 8:28 PM, dbush wrote:

 

>
DDD emulated by HHH1 emulating HHH  DDD emulated by HHH emulating HHH
[00002183] push ebp      ;          [00002183] push ebp      ;
[00002184] mov ebp,esp   ;          [00002184] mov ebp,esp   ;
[00002186] push 00002183 ; DDD      [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH      [0000218b] call 000015c3 ; HHH
                                     ### HHH aborts ###
>

 *That is dishonest*
HHH1 emulates DDD once and this matches
the first time that HHH emulates DDD.
 DDD emulated by HHH1              DDD emulated by HHH
[00002183] push ebp               [00002183] push ebp
[00002184] mov ebp,esp            [00002184] mov ebp,esp
[00002186] push 00002183 ; DDD    [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH    [0000218b] call 000015c3 ; HHH
 *THE MATCHING IS NOW USED UP*
*THE MATCHING IS NOW USED UP*
*THE MATCHING IS NOW USED UP*

Right, becuase HHH gives up when HHH1 proves that the complete emulation would halt.

 *Then DDD emulated by HHH diverges from DDD emulated by HHH1*
 DDD emulated by HHH emulating itself
[00002183] push ebp      ;
[00002184] mov ebp,esp   ;
[00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH
 

Which is the emulation done by HHH1, that shows that HHH will abort and return 0 and thus DDD halts.
Which goes against the INCORRECT assumptions that HHH used, and thus HHH was just incorrect, and you are proven to not understand how logic works,