Recent content by Preno

Does it matter whether there's a DSM label for this behaviour? What exactly is the explanatory difference between saying that a person gets off on arguing and that they suffer from, say, Argumentative Personality Disorder? Would the absence of a label for this particular form of behaviour make...

No, obviously any subtheory of an incomplete theory is also incomplete. Also, ZFC without the axiom of infinity is basically Peano Arithmetic (or rather, replacing the Axiom of infinity by its negation yields a theory which is bi-interpretable with PA).

A theory is a set of formulas closed under \vdash. So, trivially, \Sigma \vdash A and A \in \Sigma mean the same thing for any theory \Sigma.

Well, consider the trivial case: P(A given A) is trivially one, whereas P(A and A) is trivially equal to P(A).

Yes. Carnielli seems to be a student of Newton da Costa, while Weber seems to belong to the Australasian school like Graham Priest. I removed that because my personal ignorance need not reflect on the actual state of affairs in the field of paraconsistent logic. (However, speaking as an...

Well, paraconsistent logic is primarily intended as a way of handling contradictory information. The motivation for Belnap's four-valued logic is precisely that, rather than anything having to do with paradoxes. Afaik the one major paraconsistent approach that does intend to deal with the Liar...

Those proofs strike me as a bit convoluted. The fact that all empty sets are equal is a consequence of the axiom of extensionality, which is literally the most fundamental property of sets, which says that sets are equal if and only if they have the same members. If the set A has no members and...

Universal Introduction doesn't work that way. Why on Earth would it follow from the fact that, say, P holds for the number 37 that P holds for all x? a and b are constants, not variables.

No, you're not.

Maybe it would be easier if you just put your questions concerning the proof of Godel's theorem in a single thread? Anyway, Rosser's trick is used in the proof of the Representability Theorem. Without Rosser's trick, you would get the Representability Theorem in the following form...

If there is no model of Γ, ¬Φ, then trivially each model of Γ, ¬Φ is also a model of Φ. Note that what you call "the antecedent rule" is normally called Weakening.

(3) is not part of the definition of ω-inconsistency - every inconsistent theory is trivially ω-inconsistent. Other than that, yes. No. If a formula of the form \exists x \varphi(x) is independent on T, i.e. if T \nvdash \exists x \varphi(x) and T \nvdash \neg \exists x \varphi(x), then adding...

Robinson Arithmetic (Q) or Minimal Arithmetic (R), depending on how you want to axiomatize PA.

I don't think so, other than the fact that (first-order) Peano Arithmetic has infinitely many axioms, not nine - the axiom schema of Induction is, well, an axiom schema (in fact, PA can be proved not to be finitely axiomatizable).

Yes, the proof seems to be correct. Note, however, that there is no need to distinguish two cases (if the variable y is "not already universally quantified in Tk" - meaning, presumably, if it is free in Tk - and if it is not), as the argument works just the same in both cases. Even if y is...