Re: Proof that H(D,D) meets its abort criteria --self-evident truth--

On 3/18/2024 1:04 AM, Richard Damon wrote:

On 3/17/24 10:22 PM, olcott wrote:

On 3/18/2024 12:11 AM, Richard Damon wrote:

On 3/17/24 9:42 PM, olcott wrote:

On 3/17/2024 11:39 PM, Richard Damon wrote:

On 3/17/24 9:00 PM, olcott wrote:

On 3/17/2024 10:32 PM, Richard Damon wrote:

On 3/17/24 8:14 PM, olcott wrote:

On 3/17/2024 9:35 PM, Richard Damon wrote:

On 3/17/24 4:27 PM, olcott wrote:

On 3/17/2024 12:37 PM, immibis wrote:

On 17/03/24 14:11, olcott wrote:

On 3/17/2024 12:22 AM, Richard Damon wrote:

On 3/16/24 10:04 PM, olcott wrote:

On 3/17/2024 12:00 AM, Richard Damon wrote:

On 3/16/24 9:42 PM, olcott wrote:

On 3/16/2024 11:28 PM, Richard Damon wrote:

On 3/16/24 9:13 PM, olcott wrote:

On 3/16/2024 10:57 PM, Richard Damon wrote:

On 3/16/24 7:52 PM, olcott wrote:

On 3/16/2024 9:43 PM, Richard Damon wrote:

On 3/16/24 5:50 PM, olcott wrote:

On 3/16/2024 7:21 PM, Richard Damon wrote:

On 3/16/24 8:29 AM, olcott wrote:

On 3/15/2024 11:29 PM, Richard Damon wrote:

On 3/15/24 8:45 PM, olcott wrote:

H(D,D) fails to make the required mistake of reporting on what it does not see.

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But it DOES make a mistake, because it does answer the question correctly.
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You are just PROVING you think lying is ok.
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You TOTALLY don't understand the meaning of truth.
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You are REALLY just a Pathological Liar, as you have no concept of real truth,
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The original halt status criteria has the impossible requirement
that H(D,D) must report on behavior that it does not actually see.
Requiring H to be clairvoyant is an unreasonable requirement.
*The criteria shown below eliminate the requirement of clairvoyance*
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(a) If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop running
unless aborted then
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*H correctly simulates its input D until*
Means H does a correct partial simulation of D until H correctly
matches the recursive simulation non-halting behavior pattern.
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But turning out to be impposible, doesn't make it incorrect or invalid.

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*You seems to be ridiculously disingenuous about the self-evident truth*
For every possible way that H can be encoded and D(D) calls H(D,D) either H(D,D) aborts its simulation or D(D) never stops running.
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And you are incredably stupid to not see this doesn't prove what you need it to.
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Yes, if you define H to not abort, the we get a non-haltig D(D), but H doesn't answwer.
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But, if you define H to abort, then,

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We see that you changed the subject away from:
[Proof that H(D,D) meets its abort criteria]
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Nope.

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H is an algorithm that simulates its input and correctly
determines whether or not it needs to abort this simulation.
That is all that this thread's H does.
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And what defines "Need"?
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It is the set of every implementation of its spec:
(a) H(D,D) Simulate input.
(b) Determine if it needs to stop simulating its input to prevent
the simulated D(D) from never halting.
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And thus not a specific algorithm?
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Again, HOW do you determine NEED?
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That is not an algorithmic step.
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We can only verify that in retrospect.

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Do you fully understand the spec?
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Yes, but I think not the way you do.
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To me, for H to NEED to abort its simulation, that means that when giving the input to a correct simulator, that simulator will not halt.
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Yes that is correct.

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You have just proven that H doesn't need abort its simulation and the abort decision is incorrect.

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The head games of a Troll.
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For every possible way that H can be encoded and D(D)
calls H(D,D) either H(D,D) aborts its simulation or D(D)
never stops running.
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Which prove NOTHING, as D varies with H, so no D that was built with an H that aborts its simulation has had its actual halting status tested.

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*That merely changes the wording of the same truism*
∀H ∈ TM ∀D ∈ TMD such that
H(D,D) simulates its input and
D calls H(D,D) and
H(D,D) does not abort its simulation
necessitates simulated D(D) never stops running.
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Third times and still not a charm.
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All those D still use an H that doesn't abort

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*You keep talking in circles, there are only two sets*
∀H ∈ TM ∀D ∈ TMD | (H(D,D) simulates its input and D calls H(D,D))
(1) H(D,D) does not abort its simulation then simulated D(D) never stops running.
(2) H(D,D) aborts its simulation then simulated D(D) stops running.
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And your top line says NOTHING about the Ds in set (2), since nothing showed them not to run
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but your (2) admitts that D(D) will stop running, and thus the top level H didn't need to abort its simulation.
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Do you understand that each H(D,D) must either abort or fail to abort?
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And do you understand

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Yes that is what I am asking. It seems that you don't understand
the difference between X being a member of a set and X not being
a member of a set. Very elemental set theory.

 And you seem to be trying to convientely forget that each D that you talk about is DIFFERENT, base on the H that it was designed to confound.

*You keep talking in circles, there are only two sets*
∀H ∈ TM ∀D ∈ TMD | (H(D,D) simulates its input and D calls H(D,D))
(1) H(D,D) does not abort its simulation then simulated D(D) never stops running.
(2) H(D,D) aborts its simulation then simulated D(D) stops running.
*By whatever means H(D,D) places itself in (2) then H(D,D) is correct*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer