Re: Formal systems that cannot possibly be incomplete except for unknowns and unknowable

On 5/7/2025 9:48 PM, olcott wrote:

On 5/7/2025 8:30 PM, dbush wrote:

On 5/7/2025 9:20 PM, olcott wrote:

On 5/7/2025 7:44 PM, dbush wrote:

On 5/7/2025 8:19 PM, olcott wrote:

On 5/7/2025 7:15 PM, dbush wrote:

On 5/7/2025 7:40 PM, olcott wrote:

On 5/7/2025 6:31 PM, dbush wrote:

On 5/7/2025 7:26 PM, olcott wrote:

>
When N instructions of DD are emulated by HHH
according to the rules of the x86 language then

>
The subject was "DD emulated by HHH", not "N instructions of DD emulated by HHH".
>
Changing the subject is the dishonest dodge of the strawman deception.

>
>
That you and Richard construe anything less than an
infinite number of steps of DD emulated by HHH
(according to the rules of the x86 language)
as an incorrect emulation IS MORONICALLY STUPID.
>

>
The fixed immutable code of HHH simulates a fixed number X of instructions of DD, the last of which was simulated incorrectly. Any number other than X is not what HHH simulates and is therefore irrelevant to HHH.
>
UTM simulates X+Y instruction of DD correctly and reaches a final state.
>

>
I will make it easier to understand.
>
void DDD()
{
   HHH(DDD);
   return;
}
>
Can DDD simulated by HHH reach its own "return" instruction?
>

>
Category error. There is no "can" as algorithm HHH is fixed and immutable, as is algorithm DDD.
>

>
Does there exist an HHH

>
Category error.  There is only one algorithm HHH and one algorithm DDD.
>

such that DDD emulated by
HHH according to the rules of the C programming language
where the DDD element of the infinite set of HHH/DDD
pairs reaches its own "return" instruction?

>
Changing the input is not allowed.
>

>
It is like I am asking you is there a positive
number that is less than zero? You don't have
to check the positive numbers one-at-a-time.
>
We can know with complete certainty that no DDD
simulated by any HHH can possibly reach its own
"return" instruction.

>
Changing the input is not allowed.
>

>

Algorithm HHH *does not* simulate algorithm DDD to the end but instead aborts in violation of the x86 language.
>

>
there is no end to reach.

>
False.  See below.
>

>

Algorithm UTM *does* simulate algorithm DDD to the end.

 DDD emulated by HHH according to the rules of the
x86 language

Doesn't happen, as you have admitted on the record:
On 5/5/2025 8:24 AM, dbush wrote:
 > On 5/4/2025 11:03 PM, dbush wrote:
 >> On 5/4/2025 10:05 PM, olcott wrote:
 >>> On 5/4/2025 7:23 PM, Richard Damon wrote:
 >>>> But HHH doesn't correct emulated DD by those rules, as those rules
 >>>> do not allow HHH to stop its emulation,
 >>>
 >>> Sure they do you freaking moron...
 >>
 >> Then show where in the Intel instruction manual that the execution of
 >> any instruction other than a HLT is allowed to stop instead of
 >> executing the next instruction.
 >>
 >> Failure to do so in your next reply, or within one hour of your next
 >> post on this newsgroup, will be taken as you official on-the-record
 >> admission that there is no such allowance and that HHH does NOT
 >> correctly simulate DD.
 >
 > Let the record show that Peter Olcott made the following post in this
 > newsgroup after the above message:
 >
 > On 5/4/2025 11:04 PM, olcott wrote:
 >  > D *WOULD NEVER STOP RUNNING UNLESS*
 >  > indicates that professor Sipser was agreeing
 >  > to hypotheticals AS *NOT CHANGING THE INPUT*
 >  >
 >  > You are taking
 >  > *WOULD NEVER STOP RUNNING UNLESS*
 >  > to mean *NEVER STOPS RUNNING* that is incorrect.
 >
 > And has made no attempt after over 9 hours to show where in the Intel
 > instruction manual that execution is allowed to stop after any
 > instruction other than HLT.
 >
 > Therefore, as per the above criteria:
 >
 > LET THE RECORD SHOW
 >
 > That Peter Olcott
 >
 > Has *officially* admitted
 >
 > That DD is NOT correctly simulated by HHH