Tag: .999
Logic on Math StackExchange
- Does $(x\in A\land y\in B)$ follow from $xRy$ May 27, 2023Is the condition for a relation R to be injective $$(xRz\land yRz)\implies x=y$$ or is it necessary to add the quantifiers? As in $$\forall x, y, z[(x,y\in A\land z\in B \land xRz\land yRz)\implies x=y]$$ I have defined “being in R relation to” as: Given a binary relation $R\subseteq A\times B$ and the elements $x\in A$ […]Leonardo Orietti Del Duca
- Existence of some syntactic deduction May 27, 2023Given an $\mathcal{L}-$language, prove if it exists or not a deduction for: $$\exists x_1 \exists x_2 \neg \varphi \vdash \neg \exists x_1 \exists x_2 \varphi $$ My idea is that if it exists a syntactic deduction, by the Soundness Theorem it follows the semantic implication, so a set composed of the first formula and the […]Superdivinidad
- Texts on the logic of chess May 27, 2023I am looking for texts that discuss the logic of the game of chess. I am sure there are a few such texts out there. Such a text might formalize chess in first-order logic. I would be very grateful if someone gave me a list of texts on the logic of chess.user107952
- If a set is arbitrary, can you prove something about it by constructing an example of it? May 27, 2023Let $R$ be a relation from $A$ to $B.$ Prove that $\operatorname{Domain}\left(R\right)\times \operatorname{Range}\left(R\right)\not\subset R.$ I need to show that there is some couple $(a,b)\in \operatorname{Domain}(R)\times \operatorname{Range}(R)$ such that $(a,b)\notin R.$ It is easy to show by example that the theorem is true: A={1,2} B={3,4} R={(1,3),(2,4)} Domain(R)={1,2} Range(R)={3,4} Domain(R) x Range(R)={(1,3),(1,4),(2,3),(2,4)} In this case, the theorem […]lightyourassonfire
- Need help with fitch proof May 27, 2023I've been tasked with using the fitch proof system to do the following: Given ¬q, (¬p⇒(¬q⇒¬r)), (s∨r), (s⇒t), and (p⇒t), prove t. I'm experiencing great difficulty doing this. I've tried to do so by assuming ¬t, proving a contradiction and thereby deriving ¬¬t, and then using negation elimination to get t. I've also tried to […]JCKing87
- Does $\Gamma \vDash \alpha \iff \Gamma \vDash \beta$ implies $\alpha \iff \beta$? May 27, 2023I think it doesn't, because if $\Gamma \vDash \alpha \iff \Gamma \vDash \beta$, then, by definition: $v(\Gamma) = 1 \to v(\alpha) = 1 \iff v(\Gamma) = 1 \to v(\beta) = 1$, which is the same as: $(v(\Gamma) = 1 \to v(\alpha) = 1) \land v(\Gamma) = 1 \iff v(\beta) = 1$, and we can simplify […]Jonas
- Strategies for proving completeness for an extension of Intuitionistic Logic May 26, 2023Recently I’ve been working on axiomatizing a logic that results from adding a new operator to standard Intuitionistic Logic. I use $\sim$ for standard intuitionistic negation, and $\neg$ for the new “weak” negation operator. I have previously posted about this logic here Question about an extension of Intuitionistic Logic . In this post I use […]RW_123
- Shoenfield absoluteness lemma with just $\mathsf{ZF}$? May 26, 2023Shoenfield's absoluteless lemma asserts that if $\sigma$ is a $\Sigma_2^1$ statement (in arithmetic), then it is absolute to all inner models of $\mathsf{ZF} + \mathsf{DC}$. I am aware that this result applies to all models of $\mathsf{ZF}$. However, all proofs that I see uses the absoluteness of well-foundedness, which requires $\mathsf{DC}$. According to this answer […]Clement Yung
- AEC as categories May 26, 2023I have 3 questions on Abstract Elementary Classes. what is the necessary and sufficient condition that a category $\cal K$ is equivalent to and AEC. why is it necessary that some faithful functor $U:\cal K\to\mathbf{Set}$ preserves directed colimits if $\cal K$ is and AEC ? What are the morphisms in $\cal K$ if it is […]user122424
- Isomorphisms between ordered groups May 26, 2023Let $\mathcal{L}= \{c,R,f \}$ a language where $c$ is a constant symbol, $R$ is a binary relation symbol and $f$ is a binary function symbol. Let denote $\Bbb{R}^+:=\{r\in \Bbb{R} : r > 0\}$ and $\Bbb{Q}^+:=\{q\in \Bbb{Q} : q > 0\}$ Prove the following: $(\Bbb{R},0,Superdivinidad
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