XPost: comp.theory, sci.math, sci.logic
On 10/23/2021 12:13 PM, Richard Damon wrote:
On 10/23/21 12:59 PM, olcott wrote:
On 10/23/2021 10:46 AM, Richard Damon wrote:
On 10/23/21 9:44 AM, olcott wrote:
On 10/23/2021 4:59 AM, Richard Damon wrote:
On 10/22/21 11:58 PM, olcott wrote:
On 10/22/2021 6:12 PM, Richard Damon wrote:
On 10/22/21 9:33 AM, olcott wrote:
On 10/22/2021 7:04 AM, Richard Damon wrote:
On 10/21/21 9:02 PM, olcott wrote:
On 10/21/2021 7:48 PM, André G. Isaak wrote:
On 2021-10-21 17:22, olcott wrote:
On 10/21/2021 5:55 PM, André G. Isaak wrote:
On 2021-10-21 16:23, olcott wrote:
On 10/21/2021 5:12 PM, André G. Isaak wrote:
On 2021-10-21 15:59, olcott wrote:
On 10/21/2021 4:45 PM, André G. Isaak wrote: >>>>>>>>>>>>>>>>> On 2021-10-21 15:14, olcott wrote:
So then you are aware that we can attain logical >>>>>>>>>>>>>>>>>> certainty of the truth of some expressions of language >>>>>>>>>>>>>>>>>> entirely on the basis of the semantic meaning of these >>>>>>>>>>>>>>>>>> expressions of language?
Sure, for a relatively small and largely uninteresting >>>>>>>>>>>>>>>>> set of sentences. But that isn't part of epistemology, >>>>>>>>>>>>>>>>> which isn't concerned with the evaluation of linguistic >>>>>>>>>>>>>>>>> expressions.
André
Do you understand that a TM that never reaches its final >>>>>>>>>>>>>>>> state is a TM that never halts?
Yes, which has nothing to do with anything I posted above. >>>>>>>>>>>>>>>
How bout them Mets?
André
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
We can tell that Ĥq0 correctly determines the halt status >>>>>>>>>>>>>> of ⟨Ĥ⟩ ⟨Ĥ⟩ entirely on the basis of the meaning of the words.
"We" certainly cannot. Which words are you even referring >>>>>>>>>>>>> to here?
André
Self-evident truths: (a & b)
(a) We know that a TM that never reaches its final state >>>>>>>>>>>> never halts.
That's not a 'self-evident truth'. That's just the definition >>>>>>>>>>> of halting.
An expression of language is self-evidently true when it is >>>>>>>>>> verified as completely rue entirely on the basis of its meaning. >>>>>>>>>>
(b) When it is verified that the simulation of ⟨Ĥ⟩ applied >>>>>>>>>>>> to ⟨Ĥ⟩ never reaches it final state (whether or not its >>>>>>>>>>>> simulation is aborted) then we know that Ĥq0 correctly >>>>>>>>>>>> aborts the simulation of its input and transitions to qn. >>>>>>>>>>>
Assuming for sake of argument that (b) is correct, how does >>>>>>>>>>> that allow us to tell that "Ĥq0 correctly determines the halt >>>>>>>>>>> status of ⟨Ĥ⟩ ⟨Ĥ⟩ entirely on the basis of the meaning of the
words."?
André
If X necessitates Y and X then Y.
Except you have two DIFFERENT things you are calling X.
in (a) it is that a TM (itself) that never reaches its final state >>>>>>>>>
in (b) it is that a (partial) simulation of H^
A partial simulation is not the TM itself.
The huge flaw in your reasoning is that it would conclude that >>>>>>>> an actual infinite loop that is not infinitely simulated cannot >>>>>>>> be correctly determined to be an infinite loop.
No, that is NOT what I am saying, but you need to use a PROPER
proof that the loop actual is infinite.
// This is complete proof.
HERE: goto HERE;
The fact that an aborted simulation doesn't reach a halting state >>>>>>> is NOT
This is where you are always much much dumber than a box of rocks. >>>>>> When I say that you have attention deficit disorder this is not
meant to be an insult. It is something that you should really look >>>>>> into.
What, the fact that I actually KNOW something bothers your. Maybe
you should look into actually learning what is the truth.
We can know that an infinite simulation would never end using the
same sort of a process that we use to determine that an infinite
loop would never end. I have explained this many dozens of times.
Except that we know that IF H actually did what you said and never
aborted, then H(<H^>,<H^>) would never answer, and if H(<H^>,<H^>)
does abort top answer that H^(<H^>) does halt.
Likewise with the infinite loop.
Right, you don't get to 'prove' that you are right by looking at bad
version of your design and say that they don't work.
So in other words you are confirming that you believe that it is
impossible for a simulating halt decider to correctly determine that
an infinite loop never halts?
void Infinite_Loop()
{
HERE: goto HERE;
}
_Infinite_Loop()
[00000ab0](01) 55 push ebp
[00000ab1](02) 8bec mov ebp,esp
[00000ab3](02) ebfe jmp 00000ab3
[00000ab5](01) 5d pop ebp
[00000ab6](01) c3 ret
Size in bytes:(0007) [00000ab6]
You like that strawman. I guess it makes some sense as we aproach
Halloween.
You seem to be under the totally mistaken impression that just because something works for one case, it means that it works for all cases.
You rejected that H(P,P) can be decided by a simulating halt decider on
the basis that if the simulation never halts then the simulating halt
decider never returns its halts status and if the simulating halt
decider does return its halt status then the input halts.
A simulating halt decider correctly decides that H(P,P) never halts on
the same kind of axiomatic basis that it decides that an infinite loop
never halts. Because of this it decides both cases correctly.
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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