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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: DDD correctly emulated by H0 --- Why Lie?
Date: Sun, 23 Jun 2024 21:30:57 -0400
Organization: i2pn2 (i2pn.org)
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On 6/23/24 9:20 PM, olcott wrote:
> On 6/23/2024 8:13 PM, Richard Damon wrote:
>> On 6/23/24 9:00 PM, olcott wrote:
>>> On 6/23/2024 7:24 PM, Richard Damon wrote:
>>>> On 6/23/24 8:08 PM, olcott wrote:
>>>>> On 6/23/2024 6:44 PM, Richard Damon wrote:
>>>>>> On 6/23/24 7:34 PM, olcott wrote:
>>>>>>> On 6/23/2024 5:58 PM, Richard Damon wrote:
>>>>>>>> On 6/23/24 6:45 PM, olcott wrote:
>>>>>>>>>
>>>>>>>>> You know what the freak I was talking from prior
>>>>>>>>> discussions unless your brain is so damaged that
>>>>>>>>> you can't remember anything from one post to the next.
>>>>>>>>>
>>>>>>>>> In the case that you affirm that your brain <is>
>>>>>>>>> this damaged then I humbly apologize.
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> No, you don't know what you are talking about.
>>>>>>>>
>>>>>>> So you insist on lying about this verified fact?
>>>>>>>
>>>>>>> _DDD()
>>>>>>> [00002172] 55               push ebp
>>>>>>> [00002173] 8bec             mov ebp,esp
>>>>>>> [00002175] 6872210000       push 00002172 ; push DDD
>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
>>>>>>> [0000217f] 83c404           add esp,+04
>>>>>>> [00002182] 5d               pop ebp
>>>>>>> [00002183] c3               ret
>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>
>>>>>>> According to the semantics of the x86 programming language
>>>>>>> when DDD correctly emulated by H0 calls H0(DDD) this call
>>>>>>> cannot possibly return.
>>>>>>>
>>>>>>
>>>>>> I won't say it can't be true, but it hasn't been proven, largely 
>>>>>> because it seems you don't know how to do a formal logic proof.
>>>>>>
>>>>>
>>>>> Liar
>>>>>
>>>>
>>>> Then where is the proof?
>>>>
>>>> And were is the simulation that H0 did?
>>>>
>>>> Failure to show where you ACTUALLY PROVED it just shows you a liar.
>>>>
>>>> Remember the parts of a Formal Logic Proof:
>>>>
>>>
>>> You could disagree that 2 + 3 = 5 on this same Jackass basis.
>>> 2 + 3 = 5 ON THE FREAKING BASIS OF THE SEMANTICS OF ARITHMETIC.
>>
>> But I seen proofs that 2 + 3 = 5
>>
>> And that is done on a proof that uses the semantics of aritmetic.
>>
>> The phrase "Semantics of Arithmetic" though, is not a proof.
>>
>>>
>>> According to the semantics of the x86 programming language
>>> when DDD correctly emulated by H0 calls H0(DDD) this call
>>> cannot possibly return.
>>>
>>
>> Then try to prove it.
>>
> 
> I will not try any prove that 2 + 3 = 5, if you deny
> it then you are a liar.

And you don't need to, as it has been done.

Now, showing how 2 + 3 = 5 would help show you how to right an actual proof.

> 
> Likewise for the behavior of DDD correctly simulated
> by H0. A correct x86 emulator already proved this three
> years ago and you still try and get away with lying about it.

Nope. Just a fallacy of proof by example, which isn't a proof.

> 
> We have gotten it down to this ONLY LIARS WILL DISAGREE
> THAT MY PROOF IS CORRECT.

WHAT PROOF?

No proof, just means your statement is just a LIE.

> 
> DDD correctly emulated by H0 DOES NOT HALT.

TYPE ERROR.

Correct Simutation by H is not part of the definition of HALTING.

Just proves your ignorance of what you talk about.


> Likewise for P correctly emulated by H.

AGAIN TYPE ERROR.

Correct Simutation by H is not part of the definition of HALTING.

Just proves your ignorance of what you talk about.

> 
> typedef int (*ptr2)();
> int H(ptr2 P, ptr2 I);
> 
> int P(ptr2 x)
> {
>    int Halt_Status = H(x, x);
>    if (Halt_Status)
>      HERE: goto HERE;
>    return Halt_Status;
> }
> 
> int main()
> {
>    H(P,P);
> }
> 
> _P()
> [000020e2] 55               push ebp         ; housekeeping
> [000020e3] 8bec             mov ebp,esp      ; housekeeping
> [000020e5] 51               push ecx         ; housekeeping
> [000020e6] 8b4508           mov eax,[ebp+08] ; parameter
> [000020e9] 50               push eax         ; push parameter
> [000020ea] 8b4d08           mov ecx,[ebp+08] ; parameter
> [000020ed] 51               push ecx         ; push parameter
> [000020ee] e82ff3ffff       call 00001422    ; call H(P,P)
> [000020f3] 83c408           add esp,+08
> [000020f6] 8945fc           mov [ebp-04],eax
> [000020f9] 837dfc00         cmp dword [ebp-04],+00
> [000020fd] 7402             jz 00002101
> [000020ff] ebfe             jmp 000020ff
> [00002101] 8b45fc           mov eax,[ebp-04]
> [00002104] 8be5             mov esp,ebp
> [00002106] 5d               pop ebp
> [00002107] c3               ret
> Size in bytes:(0038) [00002107]
> 
> 
> 


And, P(P) Halts since you have indicated that H(P,P) to returns 0.

VERIFIED FACT.

H may not have been able to prove that statement, but it could prove 
that it doesn't halt, and basically made a wrong guess based on faulty 
logic.

YOU on the other hand, are just a LIAR, as you claim your H does 
something it doesn't.