Logic on Math StackExchange
- How to Substitute a term for a bound variable in a quantified formula of first order logic? June 17, 2025I am currently working through the textbook An Introduction to Mathematical Logic and Type Theory: to Truth through Proof, by Peter Andrews, and there have been occasions where I've felt like Andrews didn't have his terms quite straight, or left something a little ambiguous, but until now it hasn't been too big of an issue. […]Thomas.M
- Why can't we just use classes as models of ZF? [duplicate] June 17, 2025When it comes to constructing models of ZF, it seems the normal approach is to add an axiom to ZF that asserts the existence of a set $M$ such that $M$ is a model of ZF, then from this model, construct other models. But for some reason, I find this rather clunky both conceptually and […]William Oliver
- For any operation with suitable properties, is there a logic whose definability yields the same operation? June 17, 2025Let $O$ be some operation such that $$Def(A) \subseteq O(A), A \subseteq B \implies O(A) \subseteq O(B)$$ for $A$ transitive we have $O(A)$ transitive, etc. Perhaps assume furthermore that $$|O(A)| \leq |A|^{Niko Gruben
- Strong amalgamation property in the theory of modal algebras June 16, 2025I was told that the theory of modal algebras is strongly amalgamating (even superamalgamating). However, I have struggled to find a proof of this fact in the literature. I am not very familiar with modal logic which results in being lost in the literature. So I am assuming that someone with more experience could view […]user1868607
- Strong amalgamation property in integral domains June 16, 2025I read that the class of models of the theory of integral domains is not strongly amalgamating. Since this theory has strong amalgamation my understanding is that the following equation cannot hold: $f'[\mathbb{Q}] \cap g'[\mathbb{Q}] = (f' \circ f)[\mathbb{Z}] = (g' \circ g)[\mathbb{Z}]$ since I am instructed to try with an amalgam that uses $\mathbb{Z}$ […]user1868607
- $\Pi^1_1$ Uniformization by HYP (when total) June 16, 2025I think I'm being dumb but suppose that $P(f, g)$ for $f, g \in \omega^\omega$ is $\Pi^1_1$ and that $\forall f \exists g P(f, g)$. It should be true that for every $f$ there is a $g$ hyperarithmetic in $f$ such that $P(f, g)$ yes? Indeed, shouldn't there even be such a $g$ computable within […]Peter Gerdes
- Why a single natural deduction proof can cover possibly many rows of truth table for entailment? June 16, 2025Context: I'm learning formal logic by the free online book forall $x$ CALGARY - An Introduction to Formal Logic. Notice that they use uppercase script letters (e.g. $\mathscr{A}$) as metavariables to represent any TFL sentence, as they explain on page 64. Question: How does a single proof can cover possibly many rows in the truth […]linear_combinatori_probabi
- Use of consistency in proving every consistent theory has a model June 16, 2025I want to prove every consistent theory has a model, but actual proofs are too in to the weeds and even the proof sketches seem unmotivated (and also, I'm not sure where consistency is necessary). So I wrote my own proof sketch, especially to emphasize why consistency is necessary. Could someone check this proof sketch […]Pineapple Fish
- My plan for studying a research paper to obtain new results — is this a good approach? [closed] June 14, 2025I’ve been thinking about how to effectively study a research paper (let’s call it Paper X) in order to build on it and prove new results. Here is the plan I came up with: First, get a general understanding of the paper without diving into the proofs — just to grasp the big picture and […]Mousa Hamieh
- Using propositional logic to determine roles June 14, 2025I've encountered a difficult problem in my discrete maths class. The question is as follows: There are three people A, B and C. Each person may only have one role, and roles cannot be duplicated. Only three roles exist: Guide: Always tells the truth Deceiver: Always lies Trickster: May lie or tell the truth Each […]Jeremiah Boey
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