Re: Hypothetical possibilities

On 7/21/2024 11:47 AM, Fred. Zwarts wrote:

Op 21.jul.2024 om 16:27 schreef olcott:

On 7/21/2024 5:34 AM, Fred. Zwarts wrote:

Op 21.jul.2024 om 05:25 schreef olcott:

On 7/20/2024 3:03 PM, Alan Mackenzie wrote:

[ Followup-To: set ]
>
In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:
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[ .... ]
>

Olcott could not point to an error, but prefers to ignore it. So, I will
repeat it, until either an error is found, or olcott admits that HHH
cannot possibly simulate itself correctly.

>
This has the disadvantage of making your posts boring to read.  All but
one poster on this newsgroup KNOW that Olcott is wrong, here.
>
Continually repeating your argument won't get him to admit he's wrong.
Richard has been trying that for much longer than you have, with the
same lack of success.  Olcott's lack of capacity for abstract reasoning,
combined with his ignorance, combined with his arrogance, prevent him
learning at all.
>
May I suggest that you reconsider your strategy of endless repetition?
>
Thanks!
>

>
I have made slight changes to what I have been saying nearly every day.
This is my newest clearest way of saying it:
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void DDD()
{
   HHH(DDD);
   return;
}
>
DDD correctly simulated by any pure function HHH cannot
possibly reach its own return instruction.
>

>
>
Which proves that the simulation is not correct.
>

>
_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]
>
When you disagree with the semantics of the meaning
of the x86 machine code instructions of DDD you are
taking a break from reality.
>

 But since it is *you* who does not understand that skipping instructions is against the semantics of the meaning of the x86 language, we know who is far from reality.
Skipping the last cycle of the simulation of HHH, after which it would halt, is against the semantics of the x86 language.
 

It seems that your lack sufficient understanding of
ordinary programming.
I am talking about hypothetical possible ways that HHH could be encoded.
(a) HHH(DDD) is encoded to abort its simulation.
(b) HHH(DDD) is encoded to never abort its simulation.
Therefore (a) is correct and (b) is incorrect according to the
design requirements for HHH that it must halt.
It is also a truism that any input that must be aborted
is a non-halting input.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer