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

Op 13.jul.2024 om 14:18 schreef olcott:

On 7/13/2024 2:53 AM, Mikko wrote:

On 2024-07-12 14:32:28 +0000, Fred. Zwarts said:
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Op 12.jul.2024 om 15:25 schreef olcott:

On 7/12/2024 3:15 AM, Mikko wrote:

On 2024-07-11 14:12:15 +0000, olcott said:
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On 7/11/2024 1:28 AM, Mikko wrote:

On 2024-07-10 18:58:14 +0000, olcott said:
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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 ]

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In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:

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[ .... ]

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Proving that the simulation is incorrect. Because a correct simulation
would not abort a halting program halfway its simulation.

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Just for clarity, a correct simulation wouldn't abort a non-halting
program either, would it?  Or have I misunderstood this correctness?

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[ .... ]

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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.

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OK, thanks!
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In other words he is saying that when you do
1 step correctly you did 0 steps correctly.

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That is possible as "correctly" has different meaning when talking
about steps from when talking about simulations.
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*No that is always false*
When you did one anythings correctly then you did
more than zero anythings correctly.

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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.
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Fred was trying to get away with saying that when 1
step of DDD is correctly emulated by HHH that 0 steps
were emulated correctly.
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Olcott has a problem with the English language.
I said that when a program needs 2 cycles of simulation, it is incorrect to abort after 1 cycle and decide it is non-halting.
His problem seems to be that he thinks that skipping x86 instructions in the simulation does not change the behaviour of a program.
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There are more situations where he seems to have a problem with the English language. He thinks that everything greater than 2 equals infinity. When a program has more than two recursions, he thinks it is non-halting.
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It is very difficult to discuss with someone with such a poor understanding of the English language, because he continuously twists the meaning of words, both his own words as well as the words of his opponents.

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I think he is less harmful that way. His lack of clarity and obvious twisting
of the meaning of words reduce the risk that anyone would believe what he
tries to say.
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 *This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
 People disagree with this on the basis they they believe
that they can disagree with the x86 language. That is the
same as disagreeing with arithmetic, not allowed.
 

The real issue is that olcott thinks he can change the semantics of the x86 language at will. He thinks that the behaviour of a program does not change if it is halted halfway its execution. Therefore, he does not understand that when the simulation of a halting program is aborted before it reaches its end, it is an incorrect simulation according to the semantics of the x86 language.