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

On 12/2/24 9:46 PM, olcott wrote:

On 11/28/2024 1:48 PM, joes wrote:

Am Thu, 28 Nov 2024 11:57:23 -0600 schrieb olcott:

On 11/28/2024 11:06 AM, Richard Damon wrote:

On 11/28/24 11:43 AM, olcott wrote:

On 11/28/2024 9:47 AM, Richard Damon wrote:

On 11/28/24 10:01 AM, olcott wrote:

>

DDD emulated by any HHH cannot possibly reach its "ret" instruction
final halt state.

>
But that DDD CAN'T be emulated more than 4 instructions by ANY pure
function, as you can't emulate past the call HHH instruction.
>

You just aren't paying any attention at all or are woefully inaccurate
in your word choice. HHH1 does emulate all of DDD.
HHH1 <is> a pure function.

Strawman. We are talking about HHH.
>

>
HHH1 has identical source-code to HHH the only difference
is that DDD does not call HHH at all, thus does not call
HHH in recursive emulation.
>
*You said*
    DDD CAN'T be emulated more than 4 instructions by
    ANY pure function...
>
Sure it can! It can be emulated by pure function HHH1
>
It HHH1 an element of the set of pure functions?
YES IT IS THUS YOU ARE WRONG !!!
>
>

  Nope, becuase if HHH1 looks at the machine code of HHH, which wasn't part of its input, it isn.t a pure function, BY DEFINITION.