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On 8/4/2024 6:12 AM, Richard Damon wrote:No, that is not what the words mean. A complete emulation is one that isOn 8/3/24 11:00 PM, olcott wrote:A complete emulation is after all of the instructions have beenOn 8/3/2024 9:56 PM, Richard Damon wrote:WHy do you say it is impossible, it just takes forever,On 8/3/24 7:36 PM, olcott wrote:You already know that a complete emulation of a non-endingOn 8/3/2024 5:51 PM, Richard Damon wrote:Maybe to your mind filled with false facts, but it isn't true.On 8/3/24 6:15 PM, olcott wrote:It is self-evident and you know it. I do have fourOn 8/3/2024 5:07 PM, Richard Damon wrote:Just says lying YOU.The problem is that every one of those emulation is of a *DIFFERENT* input, so they don't prove anything together except that each one didn't go far enough.void DDD()
{
HHH(DDD);
return;
}
When each HHH correctly emulates 0 to infinity steps of
its corresponding DDD and none of them reach the "return"
halt state of DDD then even the one that emulated infinite
steps of DDD did not emulate enough steps?
You got any source for that other than yourself?
people (two with masters in CS) that attest to that.
*It is as simple as I can possibly make it*
I wonder how you think that you are not swearing yourBecause, I know I speak the truth.
allegiance to that father of lies?
Why do you not think you are lying?
Anyone that truly understands infinite recursion knowsRight, but for every other HHH, which the ones that answer are, it isn't a fact.
that DDD correctly simulated by HHH cannot possibly reach
its own "return" final state.
Surpisingly (to me) Jeff Barnett set the record straightNo, there is one, and only one definition, it is a machine that reaches its final state.
on exactly what halting means.
Note, *a machine*, not a (partial) emulation of the machine
sequence is impossible and you already acknowledged that
DDD emulated by HHH that never aborts is non-ending.
emulated. That never happens with any infinite execution.
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