Re: Bad faith and dishonesty

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).
>
As it happens, Olcott accepts anyway that Turing's conclusion is correct, so his only beef can be with an assumption or a step.
>

 Turing's conclusion *is correct within a false assumption*

Specifically, the assumption that the following requirements can be met:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<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

YOU MUST PAY ATTENTION TO ALL THE WORDS THAT I SAY.
 

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.

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.
Given that HHH meets the above requirements, the function call HHH(DDD) means that algorithm HHH is required to report if algorithm DDD halts when executed directly.  Since HHH(DDD)==0 and algorithm DDD halts when executed directly, HHH(DDD)==0 is wrong and the assumption that HHH meets the above requirements is proven false.

 int main()
{
   DDD(); // is not an input to the HHH that it calls.
}

False.  It follows that it is from the assumption that HHH meets the above requirements.