Re: Computable Functions --- finite string transformation rules --- 0 ≠1

On 2025-04-27 18:23:03 +0000, olcott said:

On 4/27/2025 4:51 AM, Mikko wrote:

On 2025-04-26 16:15:44 +0000, olcott said:
 

_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 any HHH emulates DD according to the finite
string transformation rules specified by the x86
language (the line of demarcation between correct
and incorrect emulation) no emulated DD can possibly
reach its final halt state and halt.

 There is a type error above. First DD is introduced as a proper name.
But later it is used in the phrase "no emulated DD" where the rules
of the language require a generic name.
 

 *This of this as an axiom schema*
No DD correctly emulated by any HHH can possibly
reach its final halt state. This conclusively
proves that every HHH is correct to reject its
input DD as non-halting.

That cannot be used as a schema before you specify what symbols in it are
placeholders and what replacements can be used for the placeholders.
--
Mikko