Re: Computable Functions --- finite string transformation rules --- 0 ≠1

On 4/24/2025 7:07 PM, Richard Damon wrote:

On 4/24/25 7:58 PM, olcott wrote:

On 4/24/2025 6:14 PM, Richard Damon wrote:

On 4/24/25 5:13 PM, olcott wrote:

On 4/24/2025 5:59 AM, Richard Damon wrote:

On 4/23/25 11:22 PM, polcott333 wrote:

On 4/23/2025 9:41 PM, Richard Damon wrote:

On 4/23/25 11:32 AM, olcott wrote:

On 4/23/2025 6:25 AM, joes wrote:

Am Tue, 22 Apr 2025 13:51:48 -0500 schrieb olcott:

On 4/22/2025 1:07 PM, Fred. Zwarts wrote:

Op 22.apr.2025 om 18:28 schreef olcott:

On 4/22/2025 7:57 AM, joes wrote:

Am Tue, 15 Apr 2025 15:44:06 -0500 schrieb olcott:

>

You continue to stupidly insist that int sum(int x, int y) {return x
+ y; }
returns 7 for sum(3,2) because you incorrectly understand how these
things fundamentally work.
>
It is stupidly wrong to expect HHH(DD) report on the direct
execution of DD when you are not telling it one damn thing about
this direct execution.

What else is it missing that the processor uses to execute it?
>

libx86emu <is> a correct x86 processor and does emulate its inputs
correctly.

>
The key thing here is that Olcott consistently does not understand that
HHH is given a finite string input that according to the semantics of
the x86 language specifies a halting program,

>
That is stupidly incorrect.

No, DD halts (when executed directly). HHH is not a halt decider, not even
for DD only.
>

People here stupidly assume that the outputs are not required to
correspond to the inputs.

But the direct execution of DD is computable from its description.
>

>
Not as an input to HHH.

>
But neither the "direct execution" or the "simulation by HHH" are "inputs" to HHH. What is the input is the representation of the program to be decided on.
>

When HHH computes halting for DD is is only allowed
to apply the finite string transformations specified
by the x86 language to the machine code of DD.

>
It is only ABLE to apply them.
>

>
The input to HHH(DD) does specify the recursive emulation
of DD including HHH emulating itself emulating DD when
one applies the finite string transformation rules of the
x86 language to THE INPUT to HHH(DD).

>
Yes, the input specifies FINITE recusive PARTIAL emulation, as the HHH that DD calls will emulate only a few instructions of DD and then return,
>

>
*You are technically incompetent on this point*
When the finite string transformation rules of the
x86 language are applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE
not even after an infinite number of emulated steps.

>
Sure it does, just after the point that HHH gives up on those transformation and aborts its (now incorrect) emulation of the input.
>

>
THAT IS COUNTER FACTUAL !!!
>
The directly executed DD has zero recursive invocations.
DD emulated by HHH has one recursive invocation.
>
Did you know that zero does not equal one?
>

 But the direct execution DOES have a recursiove invocation, as DD calls HHH(DD) that emulated DD, just like the directly exeucted HHH will emulate DD calling HHH(DD).
 

The call from the directly executed DD to HHH(DD)
immediately returns and DD reaches its final halt state.
The call to HHH(DD) from DD emulated by HHH causes
HHH to emulate itself emulating DD. DD cannot possibly
reach its final halt state without violating the finite
string transformation rules of the x86 language.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer