Re: Even Google AI Overview understands me now --- different execution traces have different behavior !!!

On 10/5/2024 10:22 PM, Richard Damon wrote:

On 10/5/24 10:58 PM, olcott wrote:

On 10/5/2024 9:46 PM, Richard Damon wrote:

On 10/5/24 9:43 AM, olcott wrote:

On 10/5/2024 8:38 AM, Richard Damon wrote:

On 10/5/24 9:34 AM, olcott wrote:

On 10/5/2024 8:27 AM, Richard Damon wrote:

On 10/5/24 8:21 AM, olcott wrote:

On 10/5/2024 5:58 AM, Richard Damon wrote:

On 10/4/24 9:53 PM, olcott wrote:

>
That you are unable to understand that it is easily conclusively
proven (below) that the emulated HHH does emulate its DDD correctly
is why your double-talk gibberish rebuttal fails.
>

>
Nope, the trace actually proves the opposite.
>

>
The following execution trace conclusively proves that
HHH emulated by itself does emulate the first four lines
of DDD correctly.

>
Right, and then makes the error of PRESUMEING INCORREDTLY that HHH(DDD) will not return,

>
That this is over-your-head really is not my mistake.
*DDD emulated by HHH cannot possibly return*

>
No, it is beyond YOUR head that the fact that HHH does abort its emulation means its doesn't show if the HHH that it was emulating will return on not.
>

>
This is simply over your head.
The infinite set of DDD emulated by HHH
never returns no matter what its corresponding HHH does.

 No, that is just a false statement based on you changing the meaning of the words.
 EVERY DDD that calls an HHH(DDD) that ever returns an answer will halt.
 

No DDD emulated by any corresponding HHH ever returns
and the HHH that emulates it does return an answer
corresponding to this behavior of this emulated DDD.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer