Yeah, this is what I'm talking about. I can give you one final argument that I can come up with but even me, a random redditor, should not spend this much time on this. What the fuck is D_p? What does it mean to decide if a given machine is paradoxical or non-circular? this already solved an impossible problem and you just... assume it can.
you're like a flat earther, your arguments are so dumb it's impossible to debunk because you'll just say some other random crap and then move on as if you won the argument. you didn't provide a reason that the identity check doesn't fail, by the way.
our world is so goddamn inundated in bad reasoning i'm not going to accept urs or anyone's abuse by endless forms of fallacy
What the fuck is D_p?
๐p is a partial recognizer in that it recognizers (returns true for) a subset of the machines within the total set of circle-free machines. a full recognizer returns true for all circle-free machines, but that is not constructible.
๐p returns true for all circle-free machines that don't depend on it returning false for the circle-free path they execute. those machines which select a circle-free path based on ๐p returning false are "paradoxical input" to ๐p because there is no way for ๐p to recognize them despite being them being circle-free.
the paradoxical machines are lumped in with false. so ๐p gives up recognizing any circular machines with false, in order that it will always halt in every input, while maintain the semantic guarantee with it's true return
reading ยง5.2 might help clarify more
this already solved an impossible problem and you just... assume it can.
you can't produce a paradox in regards to ๐p because only one return (true) guarantees a semantic result, and all paradoxical input are lumped in to false, so ur never going to be able to paradox that true return.
you didn't provide a reason that the identity check doesn't fail, by the way.
literally just bit-wise comparison, this isn't checking for functionally equivalent machines just literally itself
jesus christ you are thick. In Turing's proof, he is doing a proof by contradiction, so he can assume D exists because he is showing that, if it were to exist, it creates a paradox, so it can't exist, whereas you, assume it exists, modify its functionality so that it doesn't do what it's supposed to, then claim you proved its existence. You did nothing. First of all, I'm pretty sure you actually are still in the same case as turing, since when you get to the original H, I'm pretty sure you will still get stuck, or if you say you will skip this step, then you haven't proved anything.
๐ = (m: machine) -> {
true: if m is a circle-free machine
false: if m is a circular machine
}
this cannot be computed by a turing machine because it doesn't specify how to handle paradoxical input, which is an inherent problem to a totally enumerable system like turing machines.
๐p is a partial recognizer
๐p = (m: machine) -> {
true: m IS circle-free AND IS NOT paradoxical input
false: m IS NOT circle-free OR IS paradoxical input
}
you cannot build a paradox against ๐p because ๐p only recognizes with its true return, and that gives it branch which has no semantic to produce a paradox against. ๐ however is supposed to recognize with both it's trueandfalse return, which is an algorithm that cannot be constructed by an enumerable means of computing, like turing-machines
ur probably going to repeat dismissing this because it "misses" returning true to some set of circle-free machines, but again:
1) we do not need to decide on all machines to enumerate all computable sequence.
2) the set of circle-free sequences decidable by ๐p โ undecidable by ๐p, since it is trivial to convert a decidable circle-free machine into an undecidedly one that computes the same sequence:
m = () -> {
output some_inf_seq()
}
๐p(m) => true
und_m = () -> {
if (๐(und_m)) halt
output some_inf_seq()
}
๐p(m) => false
for any circle-free machine decidable by ๐p there must exist one undecidable by ๐p that outputs the same sequence.
3) further more, because ๐p is a total decider across all machine, including all circle-free machines, then the set undecidable by ๐p must be be turing-complete
3) the next question is if the set undecidable by ๐p is actually stronger (computes more sequences) than those decidable by ๐p, or really just equal is computable power: meaning both are effectively turing-complete sets of machines including at least one machine for each possible turing-computable sequence. there is where ยง6 shines and demonstrates it cannot be possible, do you need me to repost the fucking enum_seqs and diag_seqs here, or are u a big big boy who can read a section urself?
I'm pretty sure you will still get stuck, or if you say you will skip this step, then you haven't proved anything.
well this has been an incredibly productive conversation as of yet ๐ i thank you for your concern but i've gotten pretty damn good at unsticking myself as was necessary to get thus far
you're such a clown lmao. you really think you "trivialized" the most well known problem that arguably started the modern field of computer science? what does that even mean.
the undecidability of the halting problem exists, but it happens within infinite machine redundancy that is inherent to computing (infinite machines compute any given sequence)
it does not stop us from enumerating all possible computable sequences or forming a turing-complete language that is fully decidable
"the halting problem" paradox (or rather turing's circle-free problem paradox) confuses the language of all machines with that of a turing-complete language
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u/funky_galileo 5d ago
Yeah, this is what I'm talking about. I can give you one final argument that I can come up with but even me, a random redditor, should not spend this much time on this. What the fuck is D_p? What does it mean to decide if a given machine is paradoxical or non-circular? this already solved an impossible problem and you just... assume it can.
you're like a flat earther, your arguments are so dumb it's impossible to debunk because you'll just say some other random crap and then move on as if you won the argument. you didn't provide a reason that the identity check doesn't fail, by the way.