Re: DDD correctly emulated by HHH cannot possibly halt

Am Fri, 12 Jul 2024 07:28:56 -0500 schrieb olcott:

On 7/12/2024 6:15 AM, Richard Damon wrote:

On 7/11/24 11:30 PM, olcott wrote:

On 7/11/2024 10:18 PM, Richard Damon wrote:

On 7/11/24 10:28 PM, olcott wrote:

 
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.

We are only interested in a simulator that simulates ALL steps.

When 1,2,3... ∞ steps of DDD are correctly emulated by HHH it is a
lie to say that this many instructions were not correctly emulated
and you know it.

But only N instructions "correctly emulated" is NOT a CORRECT
emulaition of the instructions of DDD/HHH

I didn't limit it to N. I say 1 to infinity steps!

That IS a limit.

So, I guess NONE of them ever stop before reaching the end, if none of
them stop before that.

1,2,3... ∞ steps of DDD are correctly emulated is every HHH/DDD pair
that can possibly exist when HHH is a pure function x86 emulator.

Only the unlimited case matters.

No more dishonest shell game ruse

I still haven't understood what you think we are switching between.

That wastes weeks and weeks talking in circles.

It's rather your very low signal to noise ratio.

For each element of this infinite set of HHH/DDD pairs DDD never halts.
For every finite N number of emulated steps HHH halts.

How can a simulation of a nonterminating program halt?

This means that every HHH of this set that aborts its emulation of DDD
is correct to reject its DDD as non halting.

A simulator cannot abort.

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