Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2

Am Wed, 17 Jul 2024 13:13:11 -0500 schrieb olcott:

On 7/17/2024 12:18 PM, joes wrote:

Am Wed, 17 Jul 2024 08:43:04 -0500 schrieb olcott:

On 7/17/2024 8:35 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:02 schreef olcott:

On 7/17/2024 1:48 AM, Mikko wrote:

On 2024-07-16 15:57:04 +0000, olcott said:

 

The trace does not show that HHH returns so there is no basis to
think that HHH is a decider.

The trace shows the data of the executed program of HHH that does
halt.

It shows some of the data, not all, and in particular, not the
halting.

DDD emulated by HHH according to the semantic meaning of its x86
instructions never stop running unless aborted.

Bla bla.

You have shown that you do not understand the semantics of the x86
language.
HHH does abort and halt after N cycles,

That is counter-factual

Then HHH is not a decider.
 

When we examine the infinite set of every HHH/DDD pair such that:
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
>
DDD emulated by any pure function HHH according to the semantic
meaning of its x86 instructions never stops running unless aborted.

DDD only calls HHH, which, being a decider, halts.

I referred to every pure function HHH that can possibly exist.
In each case DDD never makes it past it fourth instruction. This means
that every HHH that halts is correct to reject its DDD as non-halting.
Not every HHH halts.

Only the one that doesn't abort after a finite number of steps doesn't
halt, and that one is not a decider.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.