Re: DDD correctly emulated by HHH cannot possibly halt

On 7/10/24 9:21 PM, olcott wrote:

On 7/10/2024 8:11 PM, Richard Damon wrote:

On 7/10/24 9:01 PM, olcott wrote:

On 7/10/2024 7:37 PM, Richard Damon wrote:

On 7/10/24 8:24 PM, olcott wrote:

On 7/10/2024 7:01 PM, Richard Damon wrote:

On 7/10/24 9:41 AM, olcott wrote:

On 7/10/2024 8:27 AM, joes wrote:

Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:

On 7/9/2024 11:01 PM, joes wrote:
  > That means that HHH doesn't return, in particular that it doesn't
  > abort.
DDD correctly emulated by any pure function HHH that correctly emulates
1 to ∞ steps of DDD can't make it past the above line of code no matter
what.

That line being the call to itself -> it can't simulate itself.
>

*DDD NEVER HALTS*

DDD ONLY calls HHH...
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
>
>

>
Nope, DDD does if HHH(DDD) returns.
>

>
You have a dead cat in your driveway does not mean that
you have a peanut butter sandwich on your front porch.
It has taken you at least 1000 messages to see that.
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.

>
WRONG, you don't seem to understand the difference between DDD and HHH's emualtion of it.
>
>

>
Would you bet your immortal soul that DDD simulated
by HHH (as provided above) would terminate normally?
>

>
That is a ambiguous statement, showing your attempt at deciet.
>

 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 its own machine address of 00002174 and halt.

And thus you stipulate that you are a LIAR.
By the semantic of the x86 programming language, the only correct simulation is a FULL simulation that proceeds to the final insttuction as that is what the x86 language defines, and thus NO HHH can correct simulate less than an infinite number of instructions of the program DDD, and in doing so spends virtually all of its time emulationg the instructions of HHH, showing that it will never halt.
When you admit to HHH not being actually a correct emulator and let it abort, and return, we can then establish that a DDD that calls such a PARTIAL emulator HHH will be finite and return even though the PARTAL emulation by that HHH mever sees it because it stops too soon,

 _DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
 Will you bet your immortal soul that any of the above
DDD of DDD/HHH pairs reach their own machine address
of 00002174 and halt?
 

Sure, because it is clearly true that EVERY DDD that calls an HHH(DDD) that returns will get to that final return, even though its HHH will never simulate its copy of it to that point, because it aborts its PARTIAL simulation and lies to itself about what it is doing (or more accurately, YOU LIED TO YOURSELF about the behavior of HHH and claimed a falsehood) and thus gets the wrong andwer.

Claiming ambiguity without specifically pointing this
out will be construed as deception thus swearing your
allegiance to the father of lies.
 

But you have already made that pledge, so yI guess you are just lost.
No ambiguity about that claim, DDD definitly gets to the return when run if HHH(DDD) returns.
The ambiguity comes up when you use wording that half implies you are looking at the (partial) simulation but also implies you are looking at the actual behavior.