Re: No one can correctly refute that simulating abort decider A(D,D) is correct

On 3/29/24 1:12 PM, olcott wrote:

On 3/29/2024 12:02 PM, Richard Damon wrote:

On 3/29/24 11:51 AM, olcott wrote:

On 3/29/2024 8:13 AM, Richard Damon wrote:

On 3/28/24 11:34 PM, olcott wrote:

On 3/28/2024 9:22 PM, Richard Damon wrote:

On 3/28/24 10:01 PM, olcott wrote:

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Right, but as you have admitted, the behavior of that machine code CHANGES depending on what is at the location of H.
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The behavior of D simulated by any H that can possibly exist
remains the same.

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Nope.
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The behavior of D CHANGES based on the H it attaches to.
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The behavior of D simulated by any H that can possibly exist
remains the same.

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Nope. Since you are using a "meaningless" term, it is just a non-truth bearer. Partial simulations do not reveal behavior, only complete simulations.
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Yes and your same reasoning says that mathematical induction can't
possibly work. No finite sequences of steps can every be extrapolated
to an in finite number of steps.
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Why do you say that?
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I guess you don't think all the Natural Numbers exist.
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You are just showing that you are using incorrect logic;
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Are you trying to get away with saying that you simply do not
"believe in" mathematical induction?

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You need to show that actual steps that PROVE your induction, which I don't think you actually know what they are.
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 You already know these steps as you have implicitly admitted when
you repeatedly said that when no H(D,D) aborts its simulation then
H(D,D) never halts.
 

Nope, since D changes behavior with a change of the definition of H.
That statement is not applicable.
The behavior of D for ANY other H has no bearing on the behavior of D with this H.
To claim otherwise is just a LIE.