Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---

On 8/8/24 11:48 PM, olcott wrote:

On 8/8/2024 10:34 PM, Richard Damon wrote:

On 8/8/24 11:03 PM, olcott wrote:

On 8/8/2024 9:52 PM, Richard Damon wrote:

On 8/8/24 9:15 AM, olcott wrote:

>
void DDD()
{
   HHH(DDD);
   return;
}
>
Each HHH of every HHH that can possibly exist definitely
*emulates zero to infinity instructions correctly* In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.
>
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
>
There is no need to show any execution trace at the x86 level
every expert in the C language sees that the emulated DDD
cannot possibly reaches its "return" instruction halt state.
>
Every rebuttal that anyone can possibly make is necessarily
erroneous because the first paragraph is a tautology.
>
>

>
Nope, it is a lie based on comfusing the behavior of DDD which is what "Halting" is.
>

>
Finally something besides
the strawman deception,
disagreeing with a tautology, or
pure ad hominem.
>
You must first agree with everything that I said above
before we can get to this last and final point that it
not actually directly referenced above.
>

>
Why do I need to agree to a LIE?
>
>

*Two key facts*
(a) The "return" instruction is the halt state of DDD.
(b) DDD correctly emulated by any HHH never reaches this state.
>

>
WRONG, as proven.
>
The SIMULATION BY HHH doesn't reach there, but DDD does,

Now you have to agree with (a).
 

Why? since you statement was proven false, the accuracy of one of the terms doesn't matter.
I guess you don't understand how logic works, you have already shown that there is a lie in your proof, and therefore it is wrong.