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.
It is using "global memory" outside its input to affect its behavior.
You are just proving you don't undetstand what you are talking about.
You clearly don't know what a "pure function" is.
Remember, you have made it clear that the code of HHH is *NOT* part of the input, and thus any accessing of it is the use of a global that wasn't part of the inputs to the funciton.