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Questions tagged [lo.logic]

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.

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I will start with an analogy. We know that $R^2$ and $R$ are isomorphic as sets, but not isomorphic as vector spaces or topological spaces. So we have different notions of isomorphisms, each notion ...
Ryder Rude's user avatar
4 votes
1 answer
122 views

Inspired by this older question: If $X\subseteq [0,1]$ such that $|X|=2^{\aleph_0}$, is there necessarily an order-preserving injection $\iota:[0,1]\to X$?
Dominic van der Zypen's user avatar
3 votes
2 answers
113 views

Let $\mathcal{N}$ denote the standard model of first-order Peano arithmetic, $\mathsf{PA}$, and let true arithmetic, $\mathsf{Th}(\mathcal{N})$, be the set of sentences in $\mathsf{L}_{\mathsf{PA}}$ ...
Opsimath's user avatar
3 votes
0 answers
79 views

This is related to the second part of this old question of mine. For $A\subseteq\mathbb{N}$ let $\mathfrak{N}_A=(\mathbb{N};0,1,+,A)$ be the expansion of Presburger arithmetic with (a predicate naming)...
Noah Schweber's user avatar
5 votes
0 answers
110 views

There are quite a lot large cardinal properties that have an ordinal parameter with bounded value. I'm curious that can we extend their range by collapsing cardinals. For example, consider the $\alpha$...
Reflecting_Ordinal's user avatar
3 votes
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Define: $X=\operatorname {Frege} (A) \iff \\ X= \{ y \mid \exists z: y= \{ a \subseteq A \mid a\text { is bijective to } z \}\}$ So $\operatorname {Frege}(X)$ is the set of all nonempty equivalence ...
Zuhair Al-Johar's user avatar
3 votes
0 answers
330 views

Assume $\Gamma$ is an infinite set composed of formulas (of finite length), and $A$ is a formula (of finite length). For example, $\Gamma=\{x,y\sqcup z,x_1,x_2,x_3,\cdots\}$, $A=(x\sqcap y)\sqcup(x\...
Florian's user avatar
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I am trying to figure out what would qualify as something like the mother of all logics. The motivation of this question comes from studying some model theory. There I came across this interesting ...
Clemens Bartholdy's user avatar
-4 votes
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Someone here bumped into the papers by Moshe Klein & Oded Maimon on Soft Logic? I try to understand whether their axioms actually exclude the zero-product property. Here are the axioms from one of ...
Ohm's user avatar
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7 votes
0 answers
292 views

The (second-order) Baire category theorem for open sets given by codes is provable in the base theory RCA$_0$ of reverse mathematics (see Simpson's SOSOA). Is the following version of the Baire ...
Sam Sanders's user avatar
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-1 votes
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81 views

The notion of an effectively axiomatised theory is based on computing abilities of a Turing machine. For instance, the wffs and proofs are required to be decidable, the theorems are required to be ...
Ryder Rude's user avatar
5 votes
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190 views

This is a reference request. Are there any results of the following kind? Assume $\mathrm{CH}$ and let $\mathcal{U}$ be a Ramsey ultrafilter. Let $c$ be a Cohen real. Then in $V[c]$, can $\mathcal{U}\...
Ekineme's user avatar
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2 answers
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If $(P,\leq)$ is a poset, an antichain is a set $A\subseteq P$ such that for all $a\neq b\in A$ we have $a\not\leq b$ and $b\not\leq a$. A chain is a subset $C\subseteq P$ such that for all $c,d\in C$ ...
Dominic van der Zypen's user avatar
1 vote
0 answers
470 views

I recently came across a recent paper presenting a proof of Zorn’s Lemma that seems very short and elementary. I found it interesting because the proof does not use transfinite induction or any set ...
Ethan Parker's user avatar
1 vote
1 answer
181 views

The language of this set theory, in addition to $\in$, contains a unary predicate $A$ for urelements. $\mathrm{Set}(x)$ abbreviates $\lnot A(x)$. The axioms (and axiom schemes) of Z, modified to allow ...
Zuhair Al-Johar's user avatar
14 votes
2 answers
643 views

I've heard it said many times times that $\boldsymbol{\Pi}^{1}_{n}$-determinacy implies $\boldsymbol{\Sigma}^{1}_{n+1}$-Lebesgue measurability (hence for instance $n$ many Woodin cardinals with a ...
Notgonna Doxxmyself's user avatar
3 votes
1 answer
735 views

The original Zermelo set theory explicitly allowed for urelements. What was the reason that led Zermelo to formulate the Axiom of Infinity in terms of the existence of a set of the kind that has an ...
Zuhair Al-Johar's user avatar
9 votes
0 answers
237 views

Consider the following theories: $T_1$: $\mathsf{ZFC+PD}$ where $\mathsf{PD}$ is stated as a schema. $T_2$: $\mathsf{ZFC+PD}$ where $\mathsf{PD}$ is a single sentence in the language of set theory. $...
n901's user avatar
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2 votes
1 answer
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Say that a forcing notion $\mathbb{P}$ is slow iff there is some $f:\mathbb{R}\rightarrow\mathbb{R}$ (in $V$) such that for every $\mathbb{P}$-name for a real, $\nu$, we have $\Vdash_\mathbb{P}\exists ...
Noah Schweber's user avatar
1 vote
1 answer
232 views

I was reading Topoi from Goldblat and noted that to calculate the disjunction of the internal logic of a category Set, we have to construct a characteristic function of the set: $$A = \{(1,1), (1,0), (...
Lost definition's user avatar
16 votes
1 answer
493 views

Below work in $\mathsf{ZFC+CH}$ for simplicity. Say that a (set) forcing notion $\mathbb{P}$ captures a map $f:\mathbb{R}\rightarrow\mathbb{R}$ iff there is some $\mathbb{P}$-name for a real $\nu$ ...
Noah Schweber's user avatar
-2 votes
0 answers
140 views

There are many famous unsolved problems in number theory that can be formulated by basic concepts. Two examples are Goldbach's conjecture: Every even natural number greater than 2 is the sum of two ...
Mohammad Ali Karami's user avatar
4 votes
0 answers
131 views

In a paper published in 1985, Shih-Ping Tung observed that an integer $m$ is nonzero if and only if $m=(2x+1)(3y+1)$ for some $x,y\in\mathbb Z$. In fact, we can write a nonzero integer $m$ as $\pm3^a(...
Zhi-Wei Sun's user avatar
  • 18.1k
2 votes
1 answer
235 views

Let ${}^\omega\omega$ denote the set of functions $f:\omega\to \omega$. For $f, g \in {}^\omega\omega$ we define $f\leq^* g$ if there is $N\in\omega$ such that $f(n)\leq g(n)$ for all $n\in \omega$ ...
Dominic van der Zypen's user avatar
6 votes
1 answer
367 views

For $n\in\omega$ and $x$ a real let $C_n^x$ be the canonical $\Pi^1_n(x)$-complete set. E.g. $C_1^x=\mathcal{O}^x$, etc. I recall seeing long ago the fact that, assuming large cardinals (precisely: ...
Noah Schweber's user avatar
5 votes
1 answer
294 views

Let $G$ be a Polish group and let $A\subseteq G$ be a subset with the Baire Property. Does it follow that for any $n\in \mathbb{N}$, the power $A^{n}$ also has the Baire Property? Of course, if $A$ is ...
Carlos Adrián's user avatar
1 vote
0 answers
176 views

I am reading Kunen's books on set theory and logic. In his approach, the metatheory is finitistic (which can be approximated in PRA). This implies that in the finitistic metatheory, one can do formal ...
Link L's user avatar
  • 235
18 votes
1 answer
639 views

Is the following (second-order) formula schema provable in ATR$_0$? Let $\varphi$ be an arithmetical formula satisfying For all $x, y\in \mathbb{R}$, we have that $x=_\mathbb{R}y$ implies $\varphi(x)...
Sam Sanders's user avatar
  • 4,792
0 votes
1 answer
276 views

The following material is quoted from A Crèche Course in Model Theory by Domenico Zambella, Section 15.3. $\mathcal{U}$ is how we denote the Monster model. For every $a\in\mathcal{U}^{x}$ and $b\in\...
centa's user avatar
  • 19
15 votes
1 answer
765 views

Working in $ZFC$, the statement "$0^\sharp$ exists" is often liberally taken to be one of many known equivalent statements. However, working in $Z_2$ or $ZFC^-$ (with collection, well-...
user116499's user avatar
21 votes
4 answers
2k views

Broadly speaking, the idea of “reverse mathematics” is to find equivalents to various standard mathematical statements over a weak base theory, in order to gauge the strength of theories (sets of ...
Gro-Tsen's user avatar
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5 votes
0 answers
96 views

One of the strongest results on the decidability of theories is Rabin's Tree Theorem. One way to state it is the following: tThe problem of deciding whether a sentence on the monadic second order (MSO)...
Numa Grinberg's user avatar
8 votes
1 answer
353 views

It seems that Jensen's proof of the consistency of CH + SH used class forcing, but the revelant properties are not clearly verified. I haven't learnt about class forcing, so I wonder whether it is ...
Ning Gan's user avatar
7 votes
1 answer
301 views

For a regular cardinal $\kappa$, a $\kappa$ tree $T$ is called special when there is a regressive function $f : T \to T$ (regressive in the tree order) so that the inverse image of every point is the ...
Monroe Eskew's user avatar
4 votes
1 answer
151 views

If $X'$ computes $\mathcal{O}^{Y}$ must $X$ compute $Y$? If not is there a function $\Gamma$ which guarantees that if $X'$ computes $\Gamma(Y)$ then $X$ computes $Y$? It is easy enough to see that ...
Peter Gerdes's user avatar
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8 votes
1 answer
333 views

In the paper "How connected is the intuitionistic continuum", D. van Dalen proves that in intuitionistic mathematics, the set $\mathbb{R} \setminus \mathbb{Q}$ is indecomposable, which means ...
Mohammad Tahmasbizadeh's user avatar
6 votes
1 answer
187 views

In Friedman and Magidor - The Number of Normal Measures, the authors use a nonstationary support iteration of posets, rather than the more customary countable or Easton support iterations: conditions $...
Jayde SM's user avatar
  • 2,205
9 votes
0 answers
136 views

There are two notions of n-ineffable. One is standard, defined in Baumgartner's paper, used by Harvey Friedman: κ is n-ineffable iff for every 2-coloring of $[\kappa]^{n+1}$, there is a stationary ...
Reflecting_Ordinal's user avatar
9 votes
1 answer
248 views

It is known that for $n\ge 1$, $\mathbf{\Delta}^1_{2n}$-Determinacy implies $\mathbf{\Pi}^1_{2n}$-determinacy, but as far as I know, this is a theorem in $\mathsf{ZFC}$. It brings the following first ...
Hanul Jeon's user avatar
  • 3,426
3 votes
1 answer
216 views

Define a set theoretic closed formula $\Phi$ internally saying that ``a given transitive set is equal to G"odel's $L_\alpha$ for some ordinal $\alpha$'', in the sense that if $X$ is transitive ...
Vladimir Kanovei's user avatar
15 votes
1 answer
872 views

For any set $X$, let $\{0,1\}^X$ be the collection of all functions $f:X\to\{0,1\}$. We make it into a vector space over the field $\mathbb{F}_2$ by endowing it with pointwise addition modulo $2$ and ...
Dominic van der Zypen's user avatar
3 votes
0 answers
208 views

The background for my question is somewhat related to this; there a very interesting paper is provided, but the setting and examples are somewhat different. I can add any necessary background or ...
Rodrigo Nicolau Almeida's user avatar
2 votes
1 answer
417 views

I should start with the following disclaimer that I know virtually no logic, sorry forgive me if my questions are ill-posed. I appreciate that all of this is probably completely obvious to a logician, ...
Selim G's user avatar
  • 3,118
16 votes
2 answers
1k views

Is there a formula $\phi$ in the language of set theory such that $$ \text{ZFC proves } \exists x \in \mathbb{R}:\text{ the set }A_x​:=\{y\in\mathbb{R}:\phi(x,y)\} \text{ is not Lebesgue measurable?} $...
Alexander's user avatar
  • 237
5 votes
1 answer
421 views

I was reading last evening a paper which talked a bit about behaviors on cones and I know for a fact that Martin's conjecture is one of the most active areas in computability and that it was settled ...
H.C Manu's user avatar
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19 votes
2 answers
1k views

I vividly remember seeing an affirmative answer to this question presented in seminar, but I can't track down a citation nor can I prove it myself, so now I'm doubting it's actually true: Working in $...
Noah Schweber's user avatar
13 votes
1 answer
524 views

Let $f:\mathbb{R}\rightarrow\omega_1$ be the "usual" surjection (send $r$ to the ordinal coded by $r$ according to some fixed reasonable coding system if such an ordinal exists, and to $0$ ...
Noah Schweber's user avatar
12 votes
0 answers
198 views

Analytic sets are projections of Borel sets, and are known to be Lebesgue measurable (in fact universally measurable). The question of whether measurability of analytic sets can be shown in some ...
Fanxin Wu's user avatar
  • 661
4 votes
0 answers
187 views

Let $\mathsf{QFL}$ be the quantifier-free fragment of first-order logic. We can recursively build a sequence of extensions of $\mathsf{QFL}$ by adding at each successor stage all quantifiers which are ...
Noah Schweber's user avatar
-1 votes
1 answer
178 views

In order to formulate my questions I need first to give a number of definitions. Also to help the reader I prove two preliminary relevant results. Given an infinite set $A$, let ${\cal P}^*(A)$ be ...
Athanassios Tzouvaras's user avatar

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