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.