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

Op 18.mrt.2024 om 21:40 schreef olcott:

On 3/18/2024 3:30 PM, immibis wrote:

On 18/03/24 21:20, olcott wrote:

On 3/18/2024 2:44 PM, Fred. Zwarts wrote:

Op 18.mrt.2024 om 18:43 schreef olcott:

On 3/18/2024 10:11 AM, Fred. Zwarts wrote:

Op 18.mrt.2024 om 15:44 schreef olcott:

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.

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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.

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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.
*By whatever means H(D,D) places itself in (2) then H(D,D) is correct*
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By repeating yourself, you run in circles.
There are three possible categories of H functions:
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1. Hah, It aborts and reports halting.
2. Han, It aborts and repeats non halting.
3. Hss does not abort, but simply simulates.
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What H(D,D) reports is off-topic for this post.
*We are only looking at this*
[Proof that H(D,D) meets its abort criteria --self-evident truth--]
*Thus H(D,D) aborts or H(D,D) fails to abort*
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Be clear in the naming. Is it Dan that is considered, or Dss? Dss must be aborted, because is does not halt, but Dan does halt and does not need to be aborted.

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*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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(a) If simulating abort decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted...
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*Therefore*
*Every element of (1) is incorrect and every element of (2) is correct*
*Pathological thinking to make them both seem incorrect is incorrect*
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So, Hss(Dss,Dss) should abort, but it does not.
and Han(Dan,Dan) should not abort, but it does.
The Hss that meets the abort criteria does not abort and the Han that does not meet its abort criteria does abort. So, both are wrong.

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Is it Dan that is considered, or Dss? Dss must be aborted, because is does not halt, but Dan does halt and does not need to be aborted.

 *This is what those naming conventions derive*
Everyone is saying that because H(D,D) did need to abort its simulation
to prevent D(D) from infinite execution that this proves that it never
needed to abort its simulation because it can rely on the fact that it
already aborted its simulation thus never needed to abort it.
 

You are almost there. If you stop naming all different H which the same name and all different D with the same name, your confusion may disappear.
Hss(Dss,Dss) should abort, but it does not.
and Han(Dan,Dan) should not abort, but it does.
The Hss that meets the abort criteria does not abort and the Han
that does not meet its abort criteria does abort. So, both are wrong.