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

On 7/16/24 10:34 AM, olcott wrote:

On 7/16/2024 2:10 AM, Mikko wrote:

On 2024-07-15 13:21:35 +0000, olcott said:
>

On 7/15/2024 2:52 AM, Mikko wrote:

On 2024-07-14 14:44:27 +0000, olcott said:
>

On 7/14/2024 3:48 AM, Mikko wrote:

On 2024-07-13 12:19:36 +0000, olcott said:
>

On 7/13/2024 2:55 AM, Mikko wrote:

On 2024-07-12 13:28:15 +0000, olcott said:
>

On 7/12/2024 3:27 AM, Mikko wrote:

On 2024-07-11 14:02:52 +0000, olcott said:
>

On 7/11/2024 1:22 AM, Mikko wrote:

On 2024-07-10 15:03:46 +0000, olcott said:
>

typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
}
>
int main()
{
   HHH(DDD);
}
>
We stipulate that the only measure of a correct emulation
is the semantics of the x86 programming language. By this
measure when 1 to ∞ steps of DDD are correctly emulated by
each pure function x86 emulator HHH (of the infinite set
of every HHH that can possibly exist) then DDD cannot
possibly reach past its own machine address of 0000216b
and halt.

>
For every instruction that the C compiler generates the x86 language
specifies an unambiguous meaning, leaving no room for "can".
>

>
then DDD cannot possibly reach past its own machine
address of 0000216b and halt.

>
As I already said, there is not room for "can". That means there is
no room for "cannot", either. The x86 semantics of the unshown code
determines unambigously what happens.
>

>
Of an infinite set behavior X exists for at least one element
or behavior X does not exist for at least one element.
Of the infinite set of HHH/DDD pairs zero DDD elements halt.

>
That is so far from the Common Language that I can't parse.
>

>
*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.

>
Here you attempt to use the same name for a constant programs and univesally
quantifed variable with a poorly specified range. That is a form of a well
known mistake called the "fallacy of equivocation".

>
I incorporated your suggestion in my paper.
DDD is a fixed constant finite string that calls its
HHH at the same fixed constant machine address.

>
That does not make sense. Which HHH does that DDD call? Which HHH
is at that fixed machine address?
>

>
HHH₁ to HHH∞ forming an infinite set of HHH/DDD pairs
>
HHH₁/DDD₁ to HHH∞/DDD∞ is another way to specify this
infinite set of HHH/DDD pairs.

>
You should not say "another way" before you have one way. What you
presented earlier is not a way as it did not make sense.

 DDD itself is a single immutable finite string have the exactly
same instructions at the exact same machine addresses.

But if it is just that string, it isn't a program, and the rest of your claims just become LIES.

 If you want at look at DDD less literally and more figuratively
then we can classify DDD as one member of a set of HHH/DDD pairs.
 There is no reason to do this because each DDD has the same behavior
and each HHH is called at the same fixed constant machine address.

FALSE, and a LIE.

 The only reason that I have DDD/HHH pairs is because deceitful
reviewers were trying to get away with saying tha some DDD have
different behavior.

No, it is that without pairing DDD to an HHH, it isn't a valid input.
You are just admitting you don't know what you are talking about.

 

It you
mean what you say now that is the way to say it. But you still must
specify avout each DDD which HHH it calls.
>

 *I have done that at least 1000 time in the last two years*

And every time it has been a LIE, since you also claim that the input is eqivalent to the input to the Turing Machine H, which means it must actually be a PROGRAM

 No DDD of the infinite set of HHH/DDD pairs where DDD
is emulated by HHH according to the semantics of the
x86 language ever reaches past its own machine address
0000216b.

No, every DDD paired with an HHH that aborts reaches that address, its just that the simulation of DDD by HHH that doesn't reach that.
You just don't know what a program is, as you have proven.

 _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]