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

On 2024-07-17 13:36:45 +0000, olcott said:

On 7/17/2024 3:03 AM, Mikko wrote:

On 2024-07-16 14:34:19 +0000, olcott said:
 

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.

 That string does not specify what the call to an address outside of the
string does and whether it returns.

 It need not do that. It specifies the address of HHH and the empirical
behavior of DDD correctly emulated by HHH shows this behavior.
 Also it is stipulated that HHH is an infinite set of x86 emulators
that correctly emulate 1,2,3...∞ instructions of DDD.

That stipulation contradicts the semantics of x86 and C languages which
require that all data is finite.
If HHH is an infinite set then so is the behaviour of DDD. Consequently,
the question of whether DDD halts is ill-posed and instead one must ask
what subset of DDD is the subset of DDD's that halt.

_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]
 *THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
DDD emulated by any pure function HHH according to the
semantic meaning of its x86 instructions never stops
running unless aborted.

That is ambigouls so agreement is incorrect.
--
Mikko