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

On 10/5/24 11:39 PM, olcott wrote:

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.
>

 That is a lie based on your misunderstanding of the meaning of the words.