Re: The philosophy of computation reformulates existing ideas on a new basis --- INFALLIBLY CORRECT REASONING

Am Thu, 07 Nov 2024 21:54:05 -0600 schrieb olcott:

On 11/7/2024 9:10 PM, Richard Damon wrote:

On 11/7/24 11:39 AM, olcott wrote:

On 11/7/2024 3:56 AM, Mikko wrote:

On 2024-11-06 15:26:06 +0000, olcott said:

On 11/6/2024 8:39 AM, Mikko wrote:

On 2024-11-05 13:18:43 +0000, olcott said:

On 11/5/2024 3:01 AM, Mikko wrote:

On 2024-11-03 15:13:56 +0000, olcott said:

On 11/3/2024 7:04 AM, Mikko wrote:

On 2024-11-02 12:24:29 +0000, olcott said:
>

HHH does compute the mapping from its input DDD to the actual
behavior that DDD specifies and this DOES INCLUDE HHH
emulating itself emulating DDD.

Yes but not the particular mapping required by the halting
problem.

Yes it is the particular mapping required by the halting
problem.
The exact same process occurs in the Linz proof.

The halting probelm requires that every halt decider terminates.
If HHH(DDD) terminates so does DDD. The halting problmen requires
that if DDD terminates then HHH(DDD) accepts as halting.

No that is false.
The measure is whether a C function can possibly reach its
"return" instruction final state.

Not in the original problem but the question whether a particular
strictly C function will ever reach its return instruction is
equally hard.

It has always been about whether or not a finite string input
specifies a computation that reaches its final state.

Not really. The original problem was not a halting problem but
Turing's

Exactly. The actual Halting Problem was called that by Davis in 1952.
Not the same as Turing proof.
has always been about whether or not a finite string input specifies a
computation that reaches its final state.

No, it has always been about trying to make a computation that given a
finite string representation of a program and input, decide if the
program will halt on that input.

It has never ever been about anything other than the actual behavior
that this finite string specifies. You are not stupid or ignorant about
this your knowledge and intelligence has seemed pretty good. What you
and others are is indoctrinated.
>

It should be noted that the problem STARTS with a program, which gets
represented with a finite string,

No that it incorrect. It never starts with a program. A TM cannot handle
another TM as its input. It starts with an encoding that has associated
semantics.

Silly distinction that buys you nothing.

and that string might be different for different deciders, as the
problem doesn't define a specific encoding method.
Your insistance that the problem starts with a finite-string just shows
your ignorance.

It is much dumber to think that a TM takes another actual TM as input.
It is common knowledge that this is not the case.

It is common knowledge that nobody is giving actual(?) TMs as input.

Try to show a reliable source that defines it as the string is the
DEFINITION of what is being asked about, as opposed to being a
representation of the program being asked about.

It is the semantics that the string specifies that is being asked about.

DDD specifies a non-halting computation to HHH because DDD calls HHH
in recursive simulation.

No, because the HHH that DDD calls is programmed to break that
recursive simulation, and thus make the results finite.

Now you are back to stupidly saying that DDD emulated by HHH reaches its
final halt state because it is aborted.

You are confusing your simulation levels here.
Not because itself is aborted, but because the HHH that it calls aborts.

You know that DDD emulated by HHH cannot possibly reach its own final
state (whether HHH ever aborts or not) and seem to believe that this is
irrelevant.

When HHH aborts, it halts and returns.

If you change HHH to not abort, then DDD does become non-halting, but

The infinite set of each HHH that emulates DDD (that aborts at some
point or not) is not above your educational or intellectual capacity.

The selfreference of HHH seems to be above your intellectual capacity.

HHH doesn't give the right answer. That is a DIFFERENT HHH, and thus a
DIFFERENT DDD (as DDD to be a program includes ALL the code it uses, so
it includes the code of HHH, which you changed)

*We are not even talking about HHH giving the right answer yet*
(a) DDD emulated by every HHH that aborts at some point
     or not never reaches its final state.

You mean, if DDD called a fixed simulator that didn’t change along with
the one simulating DDD.

(b) This means that the right answer would be that DDD emulated
     by HHH does not halt.

This means that HHH aborts, halts and returns "DDD doesn’t halt". Then
DDD, which calls HHH, also halts, making HHH wrong.

(c) If HHH rejects DDD as non halting then HHH is correct.
(d) Can any HHH compute the mapping from its input DDD to
     the actual behavior that DDD specifies as a pure function of its
     inputs *IS THE ONLY ACTUAL REMAINING UNRESOLVED ISSUE*

Where’s the problem?

--
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.