Re: DDD correctly emulated by HHH is correctly rejected as non-halting V2

On 7/15/2024 3:17 AM, Mikko wrote:

On 2024-07-14 14:50:47 +0000, olcott said:

On 7/14/2024 5:09 AM, Mikko wrote:

On 2024-07-12 14:56:05 +0000, olcott said:

When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
>

You should use the indices here, too, e.g., "where 1 to infinity
steps of DDD₁ are correctly emulated by HHH₃" or whatever you mean.
>

DDD is the exact same fixed constant finite string that always calls
HHH at the same fixed constant machine address.

 
If the function called by DDD is not part of the input then the input
does not specify a behaviour and the question whether DDD halts is
ill-posed.
 

We don't care about whether HHH halts. We know that HHH halts or fails
to meet its design spec.

We are only seeing if DDD correctly emulated by HHH can can possibly
reach its own final state.