Hi everyone,
I may be late to my office hours tomorrow (Thursday, 9/8), as I will be dropping my daughter off for her first day of school (second grade). I should be there by 9:30.
Regards,
Prof. Reitz
Logic on Math StackExchange
- Languages with quantification over *formulas*? May 17, 2024The 4th page of this paper mentions: A first-order language that contains “substitutional quantifiers” (with formulas as substituends) in addition to ordinary quantifiers turns out to be precisely the sort of “infinitary” language that we need So, it seems to be describing a sort of extension of first-order logic where you can quantify over formulas […]NikS
- Clarification on "absolute" property May 17, 2024I read that if some property $P(x)$ defined by a formula $\phi$ is absolute for some class $M$, then $\phi(P(x)) \leftrightarrow \phi^M(P(x))$, where $\phi(P(x))$ is interpreted in $V$, and $\phi^M(P(x))$ is relativized to $M$, which is a subclass of $V$. In most texts in set theory (i.e. proof that $R$ (axiom of regularity) is consistent […]Link L
- Godel's Second Incompleteness Theorem on $L$ May 17, 2024Am an absolute beginner here. But I've read that if a sentence $A$ is a theorem of a Theory $T$, then $T[\neg A]$ is inconsistent (where $T[\neg A]$ means that $\neg A$ is added as a logical axiom to $T$. Now, if say some axiom $A$ were shown to be not provable in $ZF$ then […]Link L
- Confusion on Section 1.2 of Rosen's Discrete Math Textbook May 16, 2024So I was able to deduce based on the rule that p implies q is the same as q unless (not p) that this is same as: (not s) -> (r -> (not q)) I could use the logical equivalence (A -> B) = (A or (not B)) to get Rosen's result, but up to […]Bob Marley
- Inaccessible cardinal in standard transitive models of ZFC May 16, 2024For a standard model of ZFC $\langle N,\in \rangle$ (where $N$ is a transitive set), if $\kappa\in N$ is an inaccessible cardinal in the model, does it imply that it is an inaccessible cardinal (in general, in the universe) ? Id est, if for $\kappa\in N$, we have $\langle N,\in\rangle\models SI(\kappa)$, then $\kappa$ is an […]Mamoun Aich
- Models of a slice of coherent category May 16, 2024I am trying to get some sense of the proof of Theorem 10 in this link. https://www.math.ias.edu/%7Elurie/278xnotes/Lecture6-Completeness.pdf Immediately after it, Exercise 11 says a model of coherent category with a chosen point is equivalent to a model of the slice category. May I please ask how precisely should I build this correspondence? Even just a […]Y.X.
- The $n$th statement in a list of $100$ statements is "Exactly $n$ of the statements in this list are false". What can we conclude? May 16, 2024The $n$th statement in a list of $100$ statements is "Exactly $n$ of the statements in this list are false". For this problem, based on the screenshotted answer, would the part I highlighted be sufficient to show that exactly 99 of them are false and the 99th statement is true? My reasoning as of now: […]Bob Marley
- Induction on two variables, confused about the induction step May 15, 2024Suppose I want to show For any $n>1$, any $r$, such that whenever $r+1 \leq n$, we have $P(n,r)$. (Geometrically, we have a 'infinite' triangle in the plane) My induction goes as $\forall n >1$ (It is like fixing $n$, as in the plane $r/n$, we cover the triangle by horizontal line) Show Base: $r=1$ […]Andrew_Ren
- Interpreting Arithmetic May 15, 2024In the context of 0-1 law. I saw some papers by Shelah and others in which they claim that in order to show that 0-1 law fails in a theory, it is enough to interpret arithmetic in it. It is not clear to me, what it exactly means. I asked some people to explain it […]Alice.H
- Prove in K modal logic $\square p , \neg \square q \vdash_K \neg \square (p \rightarrow q) $ May 15, 2024I need to prove in K modal logic the following $\square p , \neg \square q \vdash_K \neg \square (p \rightarrow q)$ $1. \square p \qquad premise$ $2. \neg \square q \qquad premise$ $3. \square (p \rightarrow q) \qquad assumption$ $4. \square (p \rightarrow q) \rightarrow (\square p \rightarrow \square q) \qquad axiom\,K$ $5. \square […]codproe
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