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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement
Date: Sat, 20 Jul 2024 11:54:39 +0300
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On 2024-07-19 14:39:25 +0000, olcott said:

> On 7/19/2024 3:51 AM, Mikko wrote:
>> On 2024-07-18 13:17:22 +0000, olcott said:
>> 
>>> On 7/18/2024 2:40 AM, Mikko wrote:
>>>> On 2024-07-17 13:00:55 +0000, olcott said:
>>>> 
>>>>> On 7/17/2024 1:43 AM, Mikko wrote:
>>>>>> On 2024-07-16 14:21:28 +0000, olcott said:
>>>>>>> 
>>>>>>> When simulated input DDD stops running {if and only if}
>>>>>>> the simulation of this input DDD has been aborted this
>>>>>>> necessitates that input DDD specifies non-halting behavior
>>>>>> 
>>>>>> DDD does not stop runnig unless it is completely exeuted.
>>>>> 
>>>>> _DDD()
>>>>> [00002163] 55         push ebp      ; housekeeping
>>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>>> [00002170] 83c404     add esp,+04
>>>>> [00002173] 5d         pop ebp
>>>>> [00002174] c3         ret
>>>>> Size in bytes:(0018) [00002174]
>>>>> 
>>>>> DDD emulated by HHH according to the semantic meaning of
>>>>> its x86 instructions never stop running unless aborted.
>>>> 
>>>> You mean HHH's simulation of DDD may not termite before HHH aborts it?
>>> 
>>> When we examine the infinite set of every HHH/DDD pair such that:
>>> HHH₁ one step of DDD₁ is correctly emulated by HHH₁.
>>> HHH₂ two steps of DDD₂ are correctly emulated by HHH₂.
>>> HHH₃ three steps of DDD₃ are correctly emulated by HHH₃.
>>> ...
>>> HHH∞ The emulation of DDD∞ by HHH∞ never stops running.
>>> 
>>> When each DDD of the HHH/DDD pairs above is correctly emulated
>>> by its corresponding HHH according to the semantic meaning of its
>>> x86 instructions it CANNOT POSSIBLY reach past its own machine
>>> address 0000216b, not even by an act of God.
>> 
>> You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
>> its termination?
> 
> No I don't mean that at all that incorrectly allocates the error
> to the emulator.

Anyway you did not say that some HHHᵢ can simulate the corresponding DDDᵢ
to its termination. And each DDDᵢ does terminate, whether simulated or not.
By the semantics of Common Language DDDᵢ is DDDᵢ, not any simulation of
DDDᵢ. As DDDᵢ is, like every mathematical thing, constant in time, your when
clause is not relevant.

> No DDDᵢ correctly emulated by its corresponding HHHᵢ can
> possibly reach past it own machine address 0000216b because
> each DDDᵢ calls its corresponding HHHᵢ in recursive emulation.

DDDᵢ can, only its emulation by HHHᵢ cannot.

>> For every finite i the behaviour specified by  DDDᵢ is
>> halting.
>> 
> 
> Unless DDDᵢ correctly emulated by its corresponding HHHᵢ
> can reach its own machine address 00002174 it cannot halt.

It does. Only its emulation by HHHᵢ doesn't.

>>>> The behaviour specified by DDD, both by C semantics and by x86 semantics,
>>>> is halting if HHH returns. Otherwise HHH is not a decider.
>>> 
>>> When HHH is required to be a pure function then only one element
>>> of the above infinite set of every possible HHH/DDD is not a decider.
>> 
>> The behavour of DDDᵢ depends on what HHHᵢ does. Wheter HHHᵢ is required
>> to what it does has no evvect on the behaviour of  DDDᵢ.
>> 
>> A pair is never a decider.
>> 
> 
> One HHH element of the above set of pairs never halts
> and thus is not a decider. Every other HHH is a decider.
> 
> Not one DDD element of the infinite set of every possible
> HHH/DDD halts, thus every HHH that halts is correct to
> reject its input as non-halting.

Every DDDᵢ with a finite i halts.

-- 
Mikko