On 5/5/2025 8:59 PM, dbush wrote:Yes.On 5/5/2025 8:57 PM, olcott wrote:Not at all.On 5/5/2025 7:49 PM, dbush wrote:>>>
Which starts with the assumption that an algorithm exists that performs the following mapping:
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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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>DO COMPUTE THAT THE INPUT IS NON-HALTING>
IFF (if and only if) the mapping FROM INPUTS
IS COMPUTED.
i.e. it is found to map something other than the above function which is a contradiction.
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The above function VIOLATES COMPUTER SCIENCE.
You make no attempt to show how my claim
THAT IT VIOLATES COMPUTER SCIENCE IS INCORRECT
you simply take that same quote from a computer
science textbook as the infallible word-of-God.
All you are doing is showing that you don't understand proof by contradiction,
The mapping is well defined. If we assume that an algorithm exists that can compute the mapping, we arrive at a contradiction, i.e. "violating computer science".
That proves the assumption false, as Linz and others have proved and as you have *explicitly* admitted is correct.
None of the things you talk about can be talked about without first making that assumption. In particular, as soon as you claim that HHH is a halt decider / termination analyzer, you are making the assumption that the mapping is computable and that HHH computes it. At that point *any* contradiction that arises, no matter how outlandish, is proof that the assumption is false.
Sorry to say, but you've wasted the last 22 years on something that 15 year olds understand.
| Date | Sujet | # | Auteur | |
| 3 May 26 |  … |
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