We prove from axioms. "constitutes" is your word. An element is an x such that there exists a y such that xey. In set theory, every x is an element of...
For sets with cardinality greater than 1: It's not that sets don't have orderings. It's that sets have many orderings (though in some cases we need a ...
No, I said "possibly". No, I'm not. I'm moving to a different metaphor. Exactly. You unnecessarily change the names and symbols for the examples. I ac...
I don't want to have to spell or copy/paste those long place names every time in discussion. Let ExUx stand for "there exists a black dog in Land U" a...
I don't know what specifically MU has in mind that I said, but I have not said anything that could be correctly paraphrased as "There are not many way...
And it's not a meaningful comparison to what I said. So we'll disregard your comment about it, after I've pointed out it was not apropos. So we'll dis...
Correct. What you're asking requires that I repeat myself. To prove ExBx, the prover might end early. To prove ~ExBx, the prover cannot end early. Agr...
So? it doesn't vitiate anything I said nor show a basis for your sarcasm. The facts are the same. But the question is not what the facts are, but what...
I really don't get you. I didn't claim that I was "nice" to do that. Only that you said that the question was not "Which is easier to prove: ExBx or ~...
Yes. But that doesn't vitiate anything I've said. I don't see a basis for your sarcasm. The thread didn't start with "black dog" and went for a while ...
In set theory, 'everything' doesn't name a thing. Rather, 'everything' is used for quantification. (1) Suppose ExAy yex. ("There exists an x such that...
An existential vs its negation. I used 'black dog' only because it came into the discussion as an example. The juncture in the discussion I have recen...
In set theory, numbers are sets. 0 = the empty set 1 = {0} 2 = {0 1} etc. This is not a claim that numbers are "really" sets (whatever "really" might ...
Doghouses don't hurt, but they're not necessary. The question was "Which is easier to prove: ExBx or ~ExBx ?" The only way that question makes sense i...
Self-inclusion is not in itself paradoxical. However, three ways to derive a contradiction from a claim that there exists a set whose members are all ...
It's too many technicalities to easily summarize. Roughly speaking, primitives: 2-place operation (x y) "pairing" 2-place operation "value of the func...
https://thephilosophyforum.com/discussion/comment/535447 I'm having second thoughts about this and I might need to retract that particular argument. T...
https://thephilosophyforum.com/discussion/comment/539584 In all cases, it seems to me that, since we are concerned with finding the shortest proof, we...
Those are my questions to sort out the discussion at StackExchange. You can post it if you want, but I probably won't follow up there or here, as I ex...
First, for your questions, these still need to be clearly settled: (1) Given a recursively axiomatizable theory with finitely many axioms, and with 'l...
At least for now, I've decided not to post there. There are already too many confusions in the discussions, and I'm not up to sorting through them wit...
In the exact sense that the set of recursive sets is a proper subset of the set of recursively enumerable sets. A theory is closed under deduction. Bu...
The very first sentence in the very first reply to you in that thread: "What does it mean for a theorem to be complete or uncountable?" Please stop us...
That is incorrect. It is an easy theorem that R is recursive iff (R is recursively enumerable & ~ R is recursively enumerable). A recursive enumeratio...
That is incorrect. You have conflated "formula F is independent of axiom set S" with "formula F is an axiom in the axiom set S". Axiomatization is a r...
Here, 'computable' is to be taken as 'Turing machine computable'. "If R is recursive then R is Turing machine computable". That has a long and complic...
That's a very modest claim. That there is a recursive axiomatization of a theory T entails that it is computable whether any given string is or is not...
I am interested to see what people there say about your notion of a 'complete theorem'. I asked you previously: Do you understand the difference betwe...
It's too much work for me to try to have you make that quote understandable. If the language is uncountable, then the theory is not recursively axioma...
What do you mean by 'computable axiom'? 'recursive axiomatization' is given a rigorous definition. With a theory that is recursively axiomatizable, it...
As Miles Davis said to producer Alfred Lion, "Is that what you wanted, Alfred?" Or, just after Davis, Hancock, Carter and Williams laid down "Thisness...
(1) Start by learning basic symbolic logic, which is the first order predicate calculus. The best textbook I have found is: 'Logic: Techniques Of Form...
Given clarity in discussion about it, yes, it is interesting. It's something I have wondered about before. If I have time later, maybe I'll reply on S...
For me, mathematical logic and set theory are chicken and egg. To formalize set theory, we use the first order predicate calculus, which is a subject ...
I was going to write: "I should have stipulated that we're talking about finitely axiomatizable theories. Even if the theory is recursively axiomatiza...
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