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

On 4/28/2025 4:10 PM, olcott wrote:

On 4/28/2025 3:02 PM, dbush wrote:

On 4/28/2025 3:53 PM, olcott wrote:

On 4/28/2025 2:10 PM, dbush wrote:

On 4/28/2025 3:03 PM, olcott wrote:

On 4/28/2025 1:37 PM, dbush wrote:

On 4/28/2025 2:32 PM, olcott wrote:

On 4/28/2025 11:11 AM, dbush wrote:

On 4/28/2025 12:10 PM, olcott wrote:

On 4/28/2025 4:05 AM, Mikko wrote:

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.
>

>
I have gone over this many hundreds of times
do you not remember anything that I already said?
>
int DD()
{
   int Halt_Status = EEE(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
When each element of the set of x86 emulators
named EEE

>
Changing the input is not allowed.

>
Hypothetical possibilities numbskull.
>

>
So you're hypothesizing changing the input.
>
Changing the input, hypothetically or otherwise, is not allowed.

>
Examining the infinite set of HHH/DD pairs simultaneously
IS NOT CHANGING THE INPUT DUMBO.
>

>
When DD runs, HHH runs.  So HHH is part of the input.  So when you hypothesize changing HHH, you're hypothesizing changing the input.
>
Changing the input, hypothetically or otherwise, is not allowed.

>
>
>
HHH[0] emulates zero instructions of DD
HHH[1] emulates one  instructions of DD
HHH[n] emulates n    instructions of DD
HHH[∞] emulates ∞    instructions of DD

>
FALSE:
>
HHH[0] emulates zero instructions of DD[0]
HHH[1] emulates one  instructions of DD[1]
HHH[n] emulates n    instructions of DD[n]
HHH[∞] emulates ∞    instructions of DD[∞]
>
>

>
Each DD has the exactly same machine code

>
False, because the algorithm DD consists of the machine code of the function DD, the function HHH, and the machine code of everything that HHH calls down to the OS level.
>

 You can look at it that way.

That IS the way.

 

So when you hypothesize changing the code of the function HHH, you're hypothesizing changing the input algorithm DD.
>
Changing the input is not allowed.

  Of the freaking infinite set of every damn HHH/DD
pair that can possibly exist where DD is emulated
by HHH according to the finite string transformation
rules of the x86 language NOT A DAMN ONE OF THE EMULATED DD HALTS.

But only one of the HHH's simulates DD.  The others simulate something else because you changed the input.
Changing the input is not allowed.

 It seems that you are trying to stupidly get away with
saying that you don't know the difference between one
element of a set and all of the elements of this set.

But the set of HHHs that simulates the actual DD only has one element.

 I know that you are not that stupid, thus are ever
playing the Troll.
 

That you're unable to understand my explanations doesn't make me a troll.