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

On 3/18/24 8:53 PM, olcott wrote:

On 3/18/2024 9:44 PM, Richard Damon wrote:

On 3/18/24 2:48 PM, olcott wrote:

On 3/18/2024 4:38 PM, Fred. Zwarts wrote:

Op 18.mrt.2024 om 22:18 schreef olcott:

On 3/18/2024 4:11 PM, Fred. Zwarts wrote:

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.

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*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.
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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.

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∀H ∈ TM ∀D ∈ TMD | (H(D,D) simulates its input and D calls H(D,D))
Every H in the above set must abort its simulated D(D).
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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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Olcott does not understand that if the H in the simulated D aborts, then the simulating H should not abort

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*You are confused*
If the H in the simulated D aborts then the directly executed H did
not abort. Since the directly executed H sees one more execution
trace then the simulated H then the H in the simulated D never aborts.

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Nope, YOU are confused If the H in the simulated D aborts,

 Then a dozen square circles are on sale at Walmart right now for $10.99

Interesting, that you retort was to just blantently lie?
When presented with FACTS, you respond with LIES.
That just shows who you are.

 

then the directly executed D MUST abort, or you are agreeing that H's simulation is not correct.
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 In other words after we have been over this hundreds and hundreds of times it is still waaaayyy over your head that the executed H always
sees exactly one more execution trace than the executed H?

Do you really read your nonsense?
How does x, "the executed H" see one more execution trace than x?
That means you think that 1 + 1 = 1

 

And no, the directed executed vesion of D see no more information then the machine the simulated D represents,

 The simulated H can not see its own behavior where as its simulator
can thus proving the simulator sees one more execution trace that its simulation.

Nope, it may see one more then at the point the simulation reaches, but the actual machine that is now being simulated did EVERYTHING that it will do as soon as it was created,
You keep on makeing that mistake, confusing the simulation with what actually happens.

 

if H aborts the simulation before then, then H just doesn't get to know what happens after that.
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I will point out, you almost NEVER actually look at the direct execution of D(D), because it just proves that H isn't a correct Halt Decider.
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 H(D,D) cannot rely on the behavior of D(D) after it has already aborted
its simulation or it would never abort its simulation and D(D) would
never stop running. This means that the executed H(D,D) see non halting
behavior.
 

Right, H is in a no-win pickle. (or its programmer is). If we wait, we run into the issue that we may never answer. If we abort, we don't know what answer to give. That is why the Halting Mapping turns out to be uncomputable.

And you seem to always confuse the different levels of the simulation, as pointed out by the fact that you "flatten" the hierarchy by displaying the simulation of a simulation as the same sort of thing as just the simulation, when it isn't.
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 The first instruction of the executed H(D,D) occurs long before the
first instruction of the simulated H(D,D) could ever happen. I can't
understand how you can't get this.

The SIMULATION BY H of the first instruction of its input happens after the first instruction of H is executed, as those DO have an ordering.
But the actual execution of the first instruction of the machine represented by the input happened long before that, when that machine was first designed.
This seems to be something foreign to you. Turing Machines don't need "something" to run, they ARE the computing device, and the program all rolled into one. As soon as you think up and design the machine, its behavior of when it runs has been fully determined. It is just a matter of learning and figuring out what that behavior is.

 Maybe this is simply over your head and Mike can explain it to you.
So far Mike and even Ben has done a great job of explaining the details
that they have explained. Ben explained how my syntax for Ĥ makes sense.
 Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt

That syntax is either:
Based on the presumption that H exsits as a correct Halt Decider (which does not actually happen, as both are contradictory)
or
The comment is a specification of which branch H (as H^.H) is supposed to take to be correct. (which shows that H can not actually make the decision that makes it correct)
I am well familiar with that notation, and I think you are still lost.

 

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and if the H in the simulated D does not abort, then the simulating H must abort.
So, the H in the simulated D must behave differently from the simulating H. Since a H cannot do both, it proves that every H in the set is incorrect.
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