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first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
4
votes
Henkin semantics for second-order logic
There are two ways for a Henkin model of second-order arithmetic to be nonstandard. 1: it could have a standard first-order part of $\omega$, but less than the full powerset of $\omega$ as its second …
3
votes
Are there "non-constructive" sets in second-order arithmetic?
Edit: I have expanded my original post somewhat. It's still marked as community wiki.
The question can be read in several ways, depending on the order in which the quantifiers are ordered. The answe …
6
votes
Is there a general setting for self-reference?
You might also be interested in Graham Priest's article "The Structure of the Paradoxes of Self-Reference", Mind 103 (1994) pp. 25-34. (Journal page ; JStor) and similar work by Priest. He has a gener …
5
votes
What does it mean for a mathematical statement to be true?
The Stanford Encyclopedia of Philosophy has several articles on theories of truth, which may be helpful for getting acquainted with what is known in the area. Their top-level article is
http://plato. …
9
votes
Impredicativity
Andrej Bauer points out that predicative constructions are more explicit, and give more useful computational information, than impredicative ones. This has two further consequences that are of intere …
6
votes
Derivation rules and Godel theorem
There are several issues here. The easiest one to resolve is the "contradiction" with Gödel's incompleteness theorem. This theorem does not concern arbitrary formal systems in your sense, but only arb …
4
votes
Accepted
Lowering order of theory
The answer to your question is that it depends on what semantics you want to use for higher-order logic.
If you use full higher-order semantics, then you cannot reduce your theory to a first-order t …
4
votes
"local variables" in first-order formulas
One elegant solution is to apply the following theorem from Enderton's logic book (Theorem 24I):
Let $\phi$ be a formula, $t$ a term, and $x$ a variable. There is a formula $\phi'$ such that $\vda …
12
votes
Accepted
Prenex normal form vs. quantifier rank
If you begin with $\psi \equiv (\exists x)R(x) \land (\exists x)(\lnot R(x)) \land (\forall y)S(y)$, the usual prenex form will have depth 3, while the original formula has depth 1. So you are asking …
22
votes
Accepted
Second-order term in first-order logic?
I think that the spirit of this question, combined with the clarifications in comments, is:
What is it that makes first-order logic "first order"?
Unfortunately, the terms "first order" and "se …
6
votes
What can be achieved by liberalizing induction for $RCA_0$?
There is at least one way in which adding induction for arithmetical formulas does "add sets" in the context of Reverse Mathematics: induction is sometimes required to show that certain "finite" sets …
3
votes
Accepted
A question on the provability predicate of Q
Robinson arithmetic is $\Sigma^0_1$ complete - every true $\Sigma^0_1$ sentence is provable in Q. If you are asking whether $Q \vdash A$ implies $Q \vdash \text{Pr}(A)$, that completeness results says …
4
votes
Existential instantiation in Hilbert-style deduction systems
For comparison, Enderton's textbook uses a Hilbert-style system. He derives EI in a form that is essentially what Emil Jeřábek calls Lemma 1', but as a metatheorem:
(EI) If $\Gamma, \phi(c) \vdash …
7
votes
First-order vs second-order provability
I assume that you mean the second-order system with both second-order induction and the full second-order comprehension scheme. There are many "second order variations" of Peano Arithmetic, with diffe …
8
votes
Accepted
Uniform solutions to Post's problem for axiomatizable theories
(Note: this has been rewritten to reflect the comments below).
The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.
Sta …