Re: DD correctly emulated by HHH --- Totally ignoring invalid rebuttals ---PSR---

Am Mon, 10 Mar 2025 09:10:10 -0500 schrieb olcott:

On 3/10/2025 6:04 AM, Richard Damon wrote:

On 3/9/25 11:10 PM, olcott wrote:

On 3/9/2025 8:57 PM, Richard Damon wrote:

On 3/9/25 6:32 PM, olcott wrote:

On 3/9/2025 4:43 PM, Richard Damon wrote:

On 3/9/25 3:36 PM, olcott wrote:

On 3/9/2025 2:24 PM, Richard Damon wrote:

On 3/9/25 9:25 AM, olcott wrote:

On 3/9/2025 6:17 AM, Richard Damon wrote:

On 3/8/25 10:24 PM, olcott wrote:

On 3/8/2025 9:03 PM, Richard Damon wrote:

On 3/8/25 6:30 PM, olcott wrote:

On 3/8/2025 5:01 PM, dbush wrote:

On 3/8/2025 5:47 PM, olcott wrote:

On 3/8/2025 4:26 PM, dbush wrote:

On 3/8/2025 11:41 AM, olcott wrote:

On 3/8/2025 9:01 AM, dbush wrote:

On 3/8/2025 9:09 AM, olcott wrote:

On 3/8/2025 3:06 AM, Mikko wrote:

On 2025-03-07 15:11:53 +0000, olcott said:

Sure I can understand what it does, as Halt7.c shows that the
behavior of the input is to HALT since that is what DD will
do when main calls it.

*WHEN YOU UNDERSTAND THIS THEN YOU KNOW YOU WERE WRONG* DD
correctly emulated by HHH cannot possibly reach its own "ret"
instruction and terminate normally because DD calls HHH(DD) in
recursive emulation.

But The HHH You are talking about doesn't do a correct
simulation, so this statment is not applicable.
>

_DD()
>

WHich is *NOT* a program, as it has an external reference.

Wrong, because emulaiting for "N Steps" is NOT correctly
emulation.

Correctly emulating N steps is emulating N steps correctly.

Which is only partially emulating it correctly, and only partially
correct is incorrect.

Everyone here that has sufficient technical competence can see
that for any N steps of DD correctly emulated by HHH that DD
cannot possibly reach its own final state and terminate normally.

So? As has been pointed out, since HHH can't do enough steps to get
to the actual answer, it never CORRECTLY emulated the input enough
to get the answer if it aborts.

If HHH can see the same pattern that every competent programmer sees
then HHH does not need to emulate DD more than twice to know that
HHH cannot possibly reach its own final state and terminate
normally.

Then HHH is not a decider.

The pattern that HHH sees is IDENTICAL to the pattern that HHH1 saw,
up to the point it aborts.

In other words you do not believe that HHH can see what every
competent programmer sees.

The problem is that what "Every Competent Programmer" will see what I
described, that since HHH aborts and returns 0, that DD will reach the
return.

When their only knowledge of HHH is that HHH emulates N steps of DD then
every competent programmer has consistently agreed that DD emulated by
HHH cannot possibly reach its own "return" instruction and terminate
normally.

We also know that HHH definitely returns, *even when simulating itself*.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.