Re: Liar detector: Peter Olcott

On 7/7/24 9:46 AM, olcott wrote:

On 7/7/2024 1:41 AM, Fred. Zwarts wrote:

Op 06.jul.2024 om 21:14 schreef olcott:

On 7/6/2024 1:55 PM, Fred. Zwarts wrote:

Op 06.jul.2024 om 18:30 schreef olcott:

On 7/6/2024 10:29 AM, Fred. Zwarts wrote:

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So, why do you disagree that the x86 code specifies an HHH that aborts and halts?

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Dishonest dodge of changing the subject. This is called
the strawman deception and is a favorite tactic of liars.

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Irrelevant text ignored. You talked about x86, therefore continuing to talk about x86 is not a change of subject.
I know you have difficulties to recognize the truth, so I do not feel offended, because: 'Don't assume somebody is wilfully wrong, if incompetence could be an explanation, as well.'
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If you sufficiently understand the semantics of the x86
language then you can see that the call to HHH(DDD) from
DDD simulated according to the semantics of the x86 language
cannot possibly return.

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I understand enough of it to see that it cannot possibly return, because HHH cannot possibly simulate itself correctly.

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According to the semantics of the x86 language IS IS IMPOSSIBLE
FOR DDD SIMULATED BY HHH TO RETURN AND IT IS EQUALLY IMPOSSIBLE
FOR THE HHH(DDD) CALLED BY DDD SIMULATED BY HHH TO RETURN.

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Therefore, you should agree that HHH cannot possibly simulate itself correctly.

 Correctly is measured by the semantics of the x86 language.
This specifies that when DDD is correctly simulated by HHH
calls emulated HHH(DDD) that this call cannot return.

but that only holds if HHH is some OTHER program that actually DOES that correct simulation, since the x86 language says that a correct simulation does not stop until the end of the program is reached.
If that is your HHH, then you lie that it returns the value 0 to say the input in non-halting.
If the input is the HHH that you are actually talking about, you LIE that it is a "Correct x86 simulator" because it isn't, it is only a PARTIAL x86 simulator, and the actual correct x86 simulation of DDD calling THIS HHH(DDD) will show that it halts.
So, you claim is based on a lie, you just get to decide which one it is.

 You smash a bottle on the ground. No matter how much you
want the bottle to hold water it will not hold water.

Right, but you had to have a bottle to smash, and not just a imaginary one.

 

That is what the semantics of the x86 teach you.
There is no disagreement about the semantics of the x86, if you see that it means that HHH cannot possibly reach its own 'ret' instruction, therefore, the simulation cannot possibly be correct.
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A correct simulation is what-so-ever-the Hell that the x86
machine code of HHH/DDD specifies even if this code starts
WW III. Correct is not measured by what you would like to
see or what you expect to happen. Correct is only measured
by the behavior that the code specifies.

Right, which is the non-aborted behavior of DDD, which will return if HHH(DDD) ever return.
Since that IS the behavior of your HHH, and as such it doesn't do a correct x86 simulation of its input, but only a PARTIAL simulation, we can say your claim is false, as DDD will return, and HHH will returun, and HHH's simulation that claims neither does is just wrong.

 When I say that 2 + 3 = 5 you are not free to dislike this
result and prefer or expect 2 + 3 = 7.
 

Right, and since the correct simulation of THIS DDD will return since you have said that HHH(DDD) will return 0 (incorrectly claimed as the correct answer) we can thus say that you HHH did NOT do a correct simulation since it got the wrong answer.

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I can't tell that you are ignorant or a liar and it is reaching
the point where I don't care which it is.
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