Re: Proof that DDD specifies non-halting behavior --- point by point

Am Wed, 14 Aug 2024 08:14:42 -0500 schrieb olcott:

On 8/14/2024 2:43 AM, joes wrote:

Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:

On 8/13/2024 9:29 PM, Richard Damon wrote:

On 8/13/24 8:52 PM, olcott wrote:

 

A simulation of N instructions of DDD by HHH according to the
semantics of the x86 language is necessarily correct.

Nope, it is just the correct PARTIAL emulation of the first N
instructions of DDD, and not of all of DDD,

That is what I said dufuss.

You were trying to label an incomplete/partial/aborted simulation as
correct.

When one instruction of DDD is correctly emulated then one instruction
was correctly emulated.

Only one, so not all.

A correct simulation of N instructions of DDD by HHH is sufficient
to correctly predict the behavior of an unlimited simulation.

Nope, if a HHH returns to its caller,

*Try to show exactly how DDD emulated by HHH returns to its caller*

how *HHH* returns

Changing the question is the strawman error or reasoning.

Richard was talking about HHH returning.

(the first one doesn't even have a caller)
Use the above machine language instructions to show this.

HHH simulates DDD         enter the matrix
   DDD calls HHH(DDD) Fred: could be eliminated
   HHH simulates DDD         second level
      DDD calls HHH(DDD) recursion detected
   HHH aborts, returns outside interference
   DDD halts     voila
HHH halts
 

That is the strawman error of reasoning.

This is not a misrepresentation of your position, this is mine.

DDD correctly emulated by HHH never reaches its own "return"
instruction. Show how it does or admit that I am correct.

See above. Show the error.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.