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

On 2024-07-11 14:10:24 +0000, olcott said:

On 7/11/2024 1:25 AM, Mikko wrote:

On 2024-07-10 17:53:38 +0000, olcott said:
 

On 7/10/2024 12:45 PM, Fred. Zwarts wrote:

Op 10.jul.2024 om 17:03 schreef olcott:

typedef void (*ptr)();
int HHH(ptr P);
 void DDD()
{
   HHH(DDD);
}
 int main()
{
   HHH(DDD);
}

 Unneeded complexity. It is equivalent to:
        int main()
       {
         return HHH(main);
       }
 

  Every time any HHH correctly emulates DDD it calls the
x86utm operating system to create a separate process
context with its own memory virtual registers and stack,
thus each recursively emulated DDD is a different instance.

 However, each of those instances has the same sequence of instructions
that the x86 language specifies the same operational meaning.
 

 *That is counter-factual*
When DDD is correctly emulated by HHH according to the
semantics of the x86 programming language HHH must abort
its emulation of DDD or both HHH and DDD never halt.

There is not "must" anywhere in the semantics of the programming language.

When DDD is correctly emulated by HHH1 according to the
semantics of the x86 programming language HHH1 need not
abort its emulation of DDD because HHH has already done this.

However, the program DDD is the same in both cases and therefore the
its behavioral meaning per x86 semantics is also the same.

The behavior of DDD emulated by HHH1 is identical to the
behavior of the directly executed DDD().

Which is the behaviour of DDD accordint to the semantics of x86 language.
--
Mikko