Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state? BUT ONLY that DDD

On 8/3/2024 5:07 PM, Richard Damon wrote:

 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?
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Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer