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On 6/13/2025 6:28 AM, Mikko wrote:The meaning of "self-evident" excludes all requirements ofOn 2025-06-11 14:11:32 +0000, olcott said:It is required that one have the technical competence of
On 6/11/2025 3:29 AM, Mikko wrote:No, it is not. The words "cannot possibly" are not sufficientlyOn 2025-06-10 16:10:49 +0000, olcott said:DDD correctly simulated by HHH cannot possibly reach its
On 6/10/2025 7:01 AM, Mikko wrote:Either the pattern or the recognition is incorrect.On 2025-06-09 14:46:30 +0000, olcott said:It only takes two simulations of DDD by HHH for HHH
On 6/9/2025 6:24 AM, Richard Damon wrote:Yes you are. You just express your stupidity in another way.On 6/8/25 10:50 PM, olcott wrote:void Infinite_Recursion()void DDD()So, you think a partial simulation defines behavior?
{
HHH(DDD);
return;
}
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*
*Every rebuttal to this changes the words*
Where do you get that LIE from?
{
Infinite_Recursion();
}
void Infinite_Loop()
{
HERE: goto HERE;
return;
}
I am no so stupid that I require a complete
simulation of a non-terminating input.
to correctly recognize a non-halting behavior pattern.
own "return" statement final halt state. This by itself
*is* complete proof that the input to HHH(DDD) specifies
non-halting behavior.
meaningful to prove anything. HHH does what it does and does
not what it does not. But what it can or cannot do, possiby or
otherwise?
a first year CS student that knows C to understand that
it is self-evident that the input to HHH(DDD) specifies
behavior such that DDD correctly simulated by HHH cannot
possibly reach its simulated "return" statement.
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