Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar

On 11/18/24 8:49 AM, olcott wrote:

On 11/18/2024 3:19 AM, joes wrote:

Am Sun, 17 Nov 2024 20:35:43 -0600 schrieb olcott:

On 11/17/2024 8:26 PM, Richard Damon wrote:

On 11/17/24 8:44 PM, olcott wrote:

On 11/17/2024 4:03 PM, Richard Damon wrote:

On 11/17/24 3:49 PM, olcott wrote:

On 11/17/2024 1:56 PM, Richard Damon wrote:

On 11/17/24 1:36 PM, olcott wrote:

>

I referred to every element of an infinite set of encodings of HHH.

>
Do you mean they are parameterised by the number of steps they simulate?
>

>
No I do not mean that.

 Then your arguement is based on an equivocation.
 

Whether or not DDD emulated by HHH ever reaches its
own "return" instruction final halt state has nothing
to do with any of the internal working of HHH as long
as each HHH emulates N steps of its input according
to the semantics of the x86 language.

 Except that the behavior DOES depend on if that HHH returns.
 Of course, your subjective, non-semantic property of "emulated by HHH" is just a meaningless term, so doesn't really mean anything, so your statement is just nonsense anyway.