Re: DD correctly emulated by HHH --- Totally ignoring invalid rebuttals ---PSR---

On 3/9/2025 9:28 AM, dbush wrote:

On 3/9/2025 10:26 AM, olcott wrote:

On 3/9/2025 9:11 AM, dbush wrote:

On 3/9/2025 10:08 AM, olcott wrote:

On 3/9/2025 8:50 AM, dbush wrote:

>
It's not an issue.

>
_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
>
When we assume that HHH emulates N steps of DD then
>
*DD correctly emulated by HHH cannot possibly reach*
*its own "ret" instruction and terminate normally*
*because DD calls HHH(DD) in recursive emulation*
>
I am not going to address any other point until this
point is fully understood because the other points
cannot possibly be understood until this one is totally
understood.
>
Whether or not and how it applies to the Halting
Theorem cannot possibly be understood at all until after
the above words are 100% totally and perfectly understood.
>

>
It is stipulated that a solution to the halting problem perform the following mapping:
>
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
>
I am not going to address any other point until this
point is fully understood because the other points
cannot possibly be understood until this one is totally
understood.

>
If you went to play head games you can play by yourself.
>

 In other words, you're disagreeing with a stipulative definition.
 As you yourself said:
 

You cannot possibly understand anything that I say
about that until you after you first understand this:
When we assume that HHH emulates N steps of DD then
DD correctly emulated by HHH cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer