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On 7/6/2025 12:00 PM, Richard Damon wrote:Looks like intolerance of dishonesty.On 7/6/25 11:19 AM, olcott wrote:You insistence that a non-terminating input be simulatedOn 7/6/2025 6:50 AM, Richard Damon wrote:No, it just isn't smart enough to detect that you lied in your premise.On 7/6/25 12:34 AM, olcott wrote:void DDD()On 7/5/2025 10:02 PM, Richard Damon wrote:Sure it does.On 7/5/25 10:43 PM, olcott wrote:With pathological self-reference the directlyOn 7/5/2025 7:44 PM, Richard Damon wrote:Really?On 7/5/25 12:26 PM, olcott wrote:It has never been the program represented by its inputOn 7/5/2025 8:14 AM, Richard Damon wrote:Nope.On 7/4/25 6:11 PM, olcott wrote:In order to have an honest dialogue you must payOn 7/4/2025 3:53 PM, Richard Damon wrote:Sure there is.On 7/4/25 4:43 PM, olcott wrote:The result is that there cannot possibly beOn 6/3/2025 10:02 PM, dbush wrote:And what is the CONTRADICTION?On 6/3/2025 10:58 PM, olcott wrote:Likewise with halt deciders that are required to reportOn 6/3/2025 9:46 PM, dbush wrote:Which arose because the axioms of naive set theory created a contradiction.On 6/3/2025 10:34 PM, olcott wrote:In the exact same way that there is no set of all setOn 6/3/2025 9:12 PM, dbush wrote:Excellent!Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:Yes there is no algorithm that does that
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
Let The Record Show
That Peter Olcott
Has *EXPLICITLY* admitted
That no algorithm H exists that meets the above requirements, which is precisely the theorem that the halting problem proofs prove.
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
on the behavior of directly executed Turing machines.
The result is just some things are not computable.
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
Your refusal to providee a source is your admission that you are just a liar.
Remember, The DEFINITION of a Halt Deicder is that it is to be a decider that decides if the program represented by its input will halt when run.
it has always been the behavior specified by its input.
This is the key mistake that no one noticed in 90 years.
In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever.
Sounds like the program and its representation.
executed machine will not have the same
behavior as the correctly simulated machine
specification.
{
HHH(DDD);
return;
}
*EVERY BOT FIGURES THIS OUT ON ITS OWN*
There is no way that DDD simulated by HHH (accordingAnd there is no way for HHH to correctly simulate its input and return an answer
to the semantics of the C programming language)
can possibly reach its own "return" statement final
halt state.
until non-existent completion is especially nuts because
you have been told about this dozens of times.
What the F is wrong with you?
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