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Not trying to argue here, just learn more. This Yablo's paradox seem self referential to me. The statements are making claims about a series of statements of w
by laxd 10y ago
Not trying to argue here, just learn more.
This Yablo's paradox seem self referential to me. The statements are making claims about a series of statements of which they themselves are part of. I see the statements only making claims about subsequent statements... but still.
How about, rather that banning "self references", you have a more carefull phrasing requiring all structures to be fully and independently defined before you start asking questions (or making claims) about them? (No side effects from the question please).
Does this avoid all Goedelish paradoxes?
- sillysaurus3 10y agoIt depends what you mean by a Godelish paradox. Godel proved that there are statements which are true, but unprovable. To me, that's quite a paradox, and there's no escaping it.
- brianberns 10y agoI think the point of the question is that perhaps we could escape it by banishing all self-referential statements?
- deleted 10y ago[deleted]
- laxd 10y agoBy Goedelish I mean all the stuff that are really just concretizations and corollaries of incompleteness. You can escape incompleteness with less powerfull axiomatic systems. But then you can't define the set of integers. (correct me if I'm wrong).
- laxd 10y agoHmm, maybe I just answered my own question. You can't define the set of integers without using elements of the set of integers. Like Peano's successor function. And if you have that power, you can state paradoxes. Like the above mentioned Yablo's paradox. ?
- mmarx 10y agoPresburger arithmetic[0] is powerful enough to describe the natural numbers with addition, yet still consistent and complete. On the other hand, Robinson arithmetic[1] lacks induction, but axiomatizes addition and multiplication, and that is sufficient to make it incomplete. Thus, the problem is not that you can describe the set of natural numbers, or induction, but the interaction between addition and multiplication. [0] https://en.wikipedia.org/wiki/Presburger_arithmetic https://en.wikipedia.org/wiki/Presburger_arithmetic [1] https://en.wikipedia.org/wiki/Robinson_arithmetic https://en.wikipedia.org/wiki/Robinson_arithmetic
- Animats 10y agoThere are a few ways around undecidability. One is finiteness. If you have an upper bound, or addition cycles around to zero, there are a finite number of states. Halting is decidable for deterministic machines with finite memory - eventually it must either halt or repeat a state. (An amusing practical solution to the halting problem is to run two interpreters in parallel, with one running twice as fast as the other. If their states match after starting, the program is in a loop. This trick was sometimes used in the batch computing era to catch beginner student programs that were in a loop before they wasted much CPU time. Programs with long cycles wouldn't be caught, but that was usually not the problem with beginner programs.)
- mmarx 10y agoDecidability (as a property of decision problems like the halting problem) is a different notion, though: In the context of incompleteness, an undecidable statement is a statement that cannot be proven or disproved in that system. Decidability of a decision problem is concerned with an effective procedure that, for every instance of the problem, answers yes/no. Specifically for a logical theory, decidability of this theory is deciding for any formula in the same language whether it follows from the theory as a logical consequence. Neither incompleteness nor undecidability imply the other[0]. [0] https://en.wikipedia.org/wiki/Decidability_(logic)#Relationship_with_completeness https://en.wikipedia.org/wiki/Decidability_(logic)#Relations...
- icen 10y agoYou can also give up recursive enumerability. `True Arithmetic`, which takes as axioms all of the true statements in PA is certainly complete and consistent.
- pvg 10y agoWhat is (inescapably) paradoxical about that?