Re: Bad faith and dishonesty

On 5/30/2025 3:05 PM, olcott wrote:

On 5/30/2025 2:01 PM, dbush wrote:

On 5/30/2025 3:00 PM, olcott wrote:

On 5/30/2025 1:48 PM, dbush wrote:

On 5/30/2025 2:40 PM, olcott wrote:

On 5/30/2025 1:20 PM, Richard Heathfield wrote:

On 30/05/2025 18:45, dbush wrote:

On 5/30/2025 1:27 PM, olcott wrote:

On 5/30/2025 12:06 PM, Richard Heathfield wrote:

There aren't many ways to invalidate a proof. Demonstrating that the conclusion is false is insufficient (because you now have two proofs, each of which claims that 'I'm right so you're wrong'); one must attack the reasoning or the assumptions (or both) and show how a flawed step or a flawed assumption invalidates the method (and perhaps the conclusion).
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As it happens, Olcott accepts anyway that Turing's conclusion is correct, so his only beef can be with an assumption or a step.
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Turing's conclusion *is correct within a false assumption*

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Specifically, the assumption that the following requirements can be met:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly

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Yes, that is precisely the assumption Turing makes, and he makes it explicitly, and he makes it with the express intent of showing that it cannot be true.
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YOU MUST PAY ATTENTION TO ALL THE WORDS THAT I SAY.

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Noise on the line again, I see. I must call the broadband people.
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Turing's only assumption is overturned by reductio within the proof itself, so that can't be it... which only leaves steps.
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As far as I can recall, Olcott's ramblings never go within discus- throwing distance of a potentially erroneous step.
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There is no *INPUT* D to termination analyzer H
that can possibly do the opposite of whatever
value that H returns.

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False. "DDD" is a description/specification of algorithm DDD consisting of the fixed code of the function DDD, the fixed code function HHH, and the fixed code of everything that HHH calls down to the OS level.

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HHH is not the computation Turing assumed could exist (for the sole purpose of showing that it could not). HHH is a hodgepodge of shit C and what looks like more line noise in assembly mnemonics. It is not a universal computation such as Turing envisaged:
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Let us suppose that there is such a process; that is to say, that we can invent a machine <D- which, when supplied with the S.D of any computing machine i l will test this S.D and if i l is circular will mark the S.D with the symbol "u" and if it is circle-free will mark it with " s ".
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By "the S.D. of any computing machine" he means the 'standard description' of >>>>any<<<< Turing machine.
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HHH is not that process, and thus HHH has no bearing whatsoever on the Turing proof.
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It is a verified fact that the
*input input input input input input*
*input input input input input input*
*input input input input input input*
*input input input input input input*
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to HHH(DDD)

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i.e. a description of algorithm DDD consisting of the fixed code of the function DDD, the fixed code of the function HHH, and the fixed code of everything that HHH calls down to the OS level.
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Never stops running unless HHH aborts its emulation

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In other words, if you change the input so that HHH doesn't abort.
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Changing the input is not allowed.

 I never changed the input you freaking moron.
 

You did exactly what when you hypothesized a different implementation of function HHH.  And since function HHH is part of the input, you changed the input.
Changing the input, hypothetically or otherwise, is not allowed.