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On 7/12/2024 3:15 AM, Mikko wrote:Olcott has a problem with the English language.On 2024-07-11 14:12:15 +0000, olcott said:Fred was trying to get away with saying that when 1
>On 7/11/2024 1:28 AM, Mikko wrote:>On 2024-07-10 18:58:14 +0000, olcott said:>
>On 7/10/2024 1:55 PM, Alan Mackenzie wrote:>Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:>Op 10.jul.2024 om 20:12 schreef Alan Mackenzie:>[ Followup-To: set ]>In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:>[ .... ]>Proving that the simulation is incorrect. Because a correct simulation
would not abort a halting program halfway its simulation.>Just for clarity, a correct simulation wouldn't abort a non-halting
program either, would it? Or have I misunderstood this correctness?>[ .... ]
>A non-halting program cannot be simulated correctly in a finite time.>
So, it depends whether we can call it a correct simulation, when it does
not abort. But, for some meaning of 'correct', indeed, a simulator
should not abort a non-halting program either.
OK, thanks!
>
In other words he is saying that when you do
1 step correctly you did 0 steps correctly.
That is possible as "correctly" has different meaning when talking
about steps from when talking about simulations.
>
*No that is always false*
When you did one anythings correctly then you did
more than zero anythings correctly.
If I only correcly do one thing that is not a part of my routine then
I don't do my routine correctly. If I do correctly every part of my routine
but do them in a wrong order I don't do my routine correctly.
>
step of DDD is correctly emulated by HHH that 0 steps
were emulated correctly.
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