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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: =?UTF-8?Q?Re=3A_Analysis_of_Flibble=E2=80=99s_Latest=3A_Detecting_v?=
 =?UTF-8?Q?s=2E_Simulating_Infinite_Recursion_ZFC?=
Date: Mon, 26 May 2025 21:35:02 +0200
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In-Reply-To: <1012epa$25ej1$1@dont-email.me>

Op 26.mei.2025 om 21:18 schreef olcott:
> On 5/26/2025 2:14 PM, Richard Heathfield wrote:
>> On 26/05/2025 18:29, olcott wrote:
>>> On 5/26/2025 12:25 PM, Richard Heathfield wrote:
>>>> On 26/05/2025 17:24, olcott wrote:
>>>>> On 5/26/2025 11:10 AM, Richard Heathfield wrote:
>>>>>> On 26/05/2025 16:42, olcott wrote:
>>>>>>> no
>>>>>>> C function can see its own caller.
>>>>>>
>>>>>> So because DDD calls HHH, HHH can't analyse the halting behaviour 
>>>>>> of DDD.
>>>>>>
>>>>>> Got it.
>>>>>>
>>>>>
>>>>> I didn't say that.
>>>>
>>>> Yes, you did.
>>>>
>>>> On 24/5/2025 in Message-ID <100sr6o$ppn2$3@dont-email.me> you said:
>>>>
>>>>> You are a damned liar when you say that I said
>>>>> that HHH must report on the behavior of its caller.
>>>>>
>>>>> No HHH can report on the behavior of its caller
>>>>> for the same reason that no function can report
>>>>> on the value of the square-root of a dead cat.
>>>>
>>>> Your words.
>>>>
>>>> Since DDD is HHH's caller, according to you HHH can't report on 
>>>> DDD's behaviour.
>>>
>>> HHH(DDD) does correctly report on the behavior that its
>>> input specifies.
>>
>> It can't. Mr Olcott said so. (See above.) You /do/ believe him, right?
>>
>>
> 
> In other words you are pretending to be so stupid that
> you don't know that the word *INPUT* and the word *CALLER*
> are not the exact same word?
We are not interested in the caller, because it is irrelevant. The input 
is a pointer to memory, including the code of Halt7.c, which specifies 
the abort and in this way specifies a halting program.

int main() {
    return HHH(DDD);
}

Now DDD is not the caller of HHH, but still a halting program is 
specified in the input. If HHH says something different, it did not 
correctly process the input.