Re: What it would take... People to address my points with reasoning instead of rhetoric -- RP

On 5/13/2025 9:35 PM, olcott wrote:

On 5/13/2025 8:26 PM, dbush wrote:

On 5/13/2025 9:16 PM, olcott wrote:

On 5/13/2025 8:03 PM, dbush wrote:

Nope.  Russell's Paradox was derived from the base axioms of naive set theory, proving the whole system was inconsistent.
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In contrast, there is nothing in existing computation theory that requires that a halt decider exists.

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I see you made no attempt to refute the above statement.  Unless you can show from the axioms of computation theory that the following requirements can be met, your argument has no basis:
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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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A halt decider doesn't exist
for the same reason that the set of all sets
that do not contain themselves does not exist.
*As defined both were simply wrong-headed ideas*

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There's nothing wrong-headed about wanting to know if any arbitrary algorithm X with input Y will halt when executed directly.

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Yes there is. I have proven this countless times.

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That requirements are impossible to satisfy doesn't make them wrong.  It just makes them impossible to satisfy, which is a perfectly reasonable conclusion.
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 It did with Russell's Paradox.
ZFC rejected the whole foundation upon which
RP was built.
 ZFC did not solve some other Russell's Paradox
it rejected the whole idea of RP as nonsense.
 

Unless you can show from the axioms of computation theory that the following requirements can be met, your argument has no basis:
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