Tag: OpenLab8
Logic on Math StackExchange
- Constructive proof of compactness theorem for countable propositional languages March 30, 2023Let $\mathcal{L}$ be a countable propositional language and let $\Gamma$ be a set of propositions of $\mathcal{L}$ (i.e. $\Gamma \subseteq \text{Prop}(\mathcal{L})$). Definition: $\Gamma$ is satisfiable if there exists an evaluation function $V:\text{Prop}(\mathcal{L}) \to \{\bot,\top\}$ such that $V(C)=\top$ for every $C \in \Gamma$ ($\{\bot,\top\}$ is the 2-elements boolean algebra). Compactness theorem for countable languages: if every […]effezeta
- Does Los's theorem hold for the filter $\{I\}$? March 30, 2023We know that $\{I\}$ is a filter over $I$. I'd like to show that $Th(M^I/ F) = Th(M)$ as a consequence of Los's theorem. Now, Los's theorem does not work in general for filters. See: What can we say if we have a filter instead of an ultrafilter in Los's theorem? Does it work for […]user1868607
- How can one show that PA is equal to this? March 30, 2023In some class note, I have seen the alternative definition of PA which is given as following: $$PA=\cup I \Sigma_n=\cup I\Pi_n$$ , while $I \Phi $ is Q equipped with the $\Phi$ induction axioms, given some class of formulas, $\Phi,$ given as below: $$\bigl\{ A(0) \implies (\forall x)(A(S(x)) \implies (\forall x)A(x): A \in \Phi \bigl\} […]io13331
- What is the justfication for splitting up statements and quantifiers? March 30, 2023To explain the title, in proving $X \times Y = \emptyset \iff X = \emptyset \vee Y = \emptyset$, we have the following \begin{align*} X \times Y = \emptyset &\iff \forall x \forall y \left((x,y) \notin X \times Y\right)\\ &\iff \forall x \forall y \left(x \notin X \vee y \notin Y\right) \\ &\iff \forall x […]jacob
- Is non-Noetherianness first order axiomatizable in the language of commutative unital rings? [duplicate] March 30, 2023By a standard argument* using Łoś's ultraproduct theorem (either directly or through the compactness theorem), Noetherianness is not first order axiomatizable in the language of commutative unital rings. Is the same true of non-Noetherianness? (Of course, it must at the very least not be finitely axiomatizable.) I've tried the obvious thing but have gotten nowhere. […]Rafi
- What ZF can do and Peano's axiom cannot. March 30, 2023I am interested in how much math can be done from Peano's axioms and what can't. What is there in the mathematics done with ZF that cannot be done with Peano's axioms?Hayatsu
- Specifying Calculus of Constructions (or something similar) in LF March 29, 2023Consider LF, the logical framework used to define UTT (unified theory of dependent types). The next two quotes are from here. LF is a simple type system with terms of the following forms: $$\textbf {Type}, El(A), (x:K)K', [x:K]k', f(k),$$ where the free occurences of variable $x$ in $K'$ and $k'$ are bound by the binding […]user125234
- Are “if / iff” interchangeable for “when / whenever”? March 29, 2023Let $A$ and $B$ be arbitrary sets and consider the following two statements: \begin{gather} (x\in A)\Rightarrow (x\in B)\\ (x\in A)\Leftrightarrow (x\in B) \end{gather} These two statements are usually worded as follows: If $x$ belongs to $A$, then $x$ belongs to $B$” “$x$ belongs to $A$ if and only if $x$ belongs to $B$”. My questions […]Hector
- How do Robinson arithmetics axioms prevent this model of N? March 29, 2023In the Peano arithmetic wikipedia article, a figure is shown on why the axiom of induction is necessary, without it, the set of whole dominos would be a valid representation of N In Robinson arithmetic, we have the axiom y=0 ∨ ∃x (Sx = y) instead. I don't see how this axiom prevent that situation, […]user2370139
- Changing EM blueprints in Abstract Elementary Classes March 29, 2023I'm currently reading some notes on AECs (here), and just wanted to make sure I understand the argument in Corollary 12.8, for I'm still getting used to generalized Ehrenfeucht-Mostowski models in AECs. As far as I understand, EM blueprints/templates $\Phi$ keep track of the quantifier-free sentences some indiscernible satisfies (that is, a single $n$-type for […]interregno
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