Am Fri, 25 Apr 2025 13:42:15 -0500 schrieb olcott:
On 4/25/2025 8:36 AM, Richard Damon wrote:
On 4/25/25 9:08 AM, olcott wrote:
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:
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.
No it doesn't,
The call starts simulating DD calling HHH, just like in the simulated DD.
The call from the directly executed DD returns.
The call from DD emulated by HHH to HHH(DD) (according to the finite
string transformation rules of the x86 language) CANNOT POSSIBLY RETURN.
It would return if HHH could simulate it. It is not non-halting, only
HHH descends ever deeper into the simulation.
--
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.
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