Re: Who here understands that the last paragraph is Necessarily true?

On 7/15/2024 4:03 AM, joes wrote:

Am Sun, 14 Jul 2024 22:41:24 -0500 schrieb olcott:

On 7/14/2024 9:04 PM, Richard Damon wrote:

On 7/14/24 9:27 PM, olcott wrote:

>
Any input that must be aborted to prevent the non termination of
simulating termination analyzer HHH necessarily specifies non-halting
behavior or it would never need to be aborted.

>
Excpet, as I have shown, it doesn't.
Your problem is you keep on ILEGALLY changing the input in your
argument because you have misdefined what the input is.

Specifically, the input HHH aborts.
 

The input to HHH is ALL of the memory that it would be accessed in a
correct simulation of DDD, which includes all the codd of HHH, and
thus, if you change HHH you get a different input.
>
If you want to try to claim the input is just the bytes of the function
DDD proper then you are just admitting that you are nothing more than a
lying idiot that doesn't understand the problem,

Turing machines only operate on finite strings they do not operate on
other Turing machines *dumbo*

Don't deflect. HHH as part of DDD (because it is called) needs to be
included in the input to the simulator.
Bedsides, TMs can be encoded as strings. Notwithstanding that HHH is not
and does not simulate a TM.
 

Richard insists that HHH report on the behavior of the TM that
is *not* encoded as finite string. TM's are not allowed to report
on the behavior of the computation that they are contained within.
The question is not whether or not HHH halts.
The question is does the finite string input to HHH mathematically
map to behavior that halts when DDD is correctly emulated by HHH
according to the semantics of the x86 language?
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer