Re: DDD correctly emulated by HHH is INcorrectly rejected as non-halting.

On 7/14/24 10:44 AM, olcott wrote:

On 7/14/2024 3:48 AM, Mikko wrote:

On 2024-07-13 12:19:36 +0000, olcott said:
>

On 7/13/2024 2:55 AM, Mikko wrote:

On 2024-07-12 13:28:15 +0000, olcott said:
>

On 7/12/2024 3:27 AM, Mikko wrote:

On 2024-07-11 14:02:52 +0000, olcott said:
>

On 7/11/2024 1:22 AM, Mikko wrote:

On 2024-07-10 15:03:46 +0000, olcott said:
>

typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
}
>
int main()
{
   HHH(DDD);
}
>
We stipulate that the only measure of a correct emulation
is the semantics of the x86 programming language. By this
measure when 1 to ∞ steps of DDD are correctly emulated by
each pure function x86 emulator HHH (of the infinite set
of every HHH that can possibly exist) then DDD cannot
possibly reach past its own machine address of 0000216b
and halt.

>
For every instruction that the C compiler generates the x86 language
specifies an unambiguous meaning, leaving no room for "can".
>

>
then DDD cannot possibly reach past its own machine
address of 0000216b and halt.

>
As I already said, there is not room for "can". That means there is
no room for "cannot", either. The x86 semantics of the unshown code
determines unambigously what happens.
>

>
Of an infinite set behavior X exists for at least one element
or behavior X does not exist for at least one element.
Of the infinite set of HHH/DDD pairs zero DDD elements halt.

>
That is so far from the Common Language that I can't parse.
>

>
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.

>
Here you attempt to use the same name for a constant programs and univesally
quantifed variable with a poorly specified range. That is a form of a well
known mistake called the "fallacy of equivocation".
>

 I incorporated your suggestion in my paper.
DDD is a fixed constant finite string that calls its
HHH at the same fixed constant machine address.

Which makes DDD not a computation, and thus not a valid thing to be talking about.
Your definition forces you to ADD all of memory to you input, and that break the rest of your logic.

 When we examine the infinite set of every HHH/DDD pair such that:

And every one of these is either an INVALID problem, because you DDD isn't actually a program to decide on, or it a totally different problem because the input is different in each case (since it includes all the memory)

HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
 The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.

And either

 No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.

Wrong, EVERY DDD instnace of an HHH/DDD pair that aborts its simulation and returns will reach past its own meachine address of 0000216b and halt, because "DDD" means the direct execution of that program.
You seem to mean the SIMULATION of DDD by HHH never reaches that point, but partial simulation, as what we get for every HHH that aborts it simulation, does not, by itself, indicate the halting behaivor of the input.

 Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
 

Nope. Since either the input is invalid, or every HHH gets a different input, you can't use the behavior of a DIFFERENT input to argue about this input, and partial simulation don't, by themselves (which is all you have) show non-halting.