Tag: openlab 7
Logic on Math StackExchange
- Examples of geometric theory without classical models April 24, 2024I encountered a post on locales and geometric theory here https://grossack.site/2022/05/22/locale-basics.html In about the middle of this blog, the author gives a geometric theory defining a function from $\Bbb N$ to $\Bbb R$, and asked: Do you see why this is a nontrivial theory with no classical models? I am not confident that I understand […]Y.X.
- can somone help me finish this fitch proof? April 24, 2024Premise 1: $P \to Q$ Premise 2: $\lnot P \to R$ Premise 3: $\lnot Q$ Goal: $R.$ enter image description here the book is Language, Proof and Logic The Second Edition authors are Jon Barwise and John EtchemendyAvrgDude
- Question relating isomorphisms and structures in logic April 24, 2024A question in a sheet was as follows: Let $T_n,$ $n \in\mathbb N \setminus 0,$ be the structure $⟨N; f_n)$ where $f_n(a) = \lfloor\frac{a}{n}\rfloor.$ Draw a picture of Tn for n = 1, 2, 3. For T1 I drew a number-line from 1 to infinity For T2 I drew a number-line of 0, 1, 1, […]Alice
- Derivation of the "Chain syllogism" rule in first order logic April 24, 2024I'm currently reading these: https://www.mcmp.philosophie.uni-muenchen.de/students/math/math_logic_munich.pdf lecture notes and am kind of stuck on the proof on page 58 line 4 where $$\Gamma \lnot \varphi \lnot \varphi$$ is derived using (Ass.). I don't understand how, none of the rule thus far introduced allow me to manipulated the consequent except contradiction which I can't use in this […]Fabio
- Does “false” imply “true”? [duplicate] April 23, 2024I think calling a proposition an "implicature" means something like this: If there is a proposition "p->q" and this proposition is true, then we think that p "implies" q. So if p is false and q is true, does “false” imply “true”? My own thoughts are, Premise: ((I am a cup) and (the cup is […]Display name
- Show that Proof by Contradiction rule is derivable from all instances of it with atomic conclusion April 23, 2024The problem shows up in Troelstra's Basic Proof Theory, exercise 2.3.6A. Show that in [classical logic's Natural Deduction], for the languages without ∨, ∃, all instances of ⊥c derivable from instances ⊥c with atomic conclusions. I have a solution to the problem, but I'm not sure if it's correct as it seems trivials, perhaps because […]confusedcius
- Define something based on the property of another set. April 22, 2024I have always been confused about a logic problem, which I can not form it into a general statement. Here is an example that may express my ideas. We know given a set $A$ and an equivalent relation $\sim$ on it. We can define equivalence class by $\forall x\in A, [x]=\{y,|y \sim x\}$. It seems […]Andrew_Ren
- Prove $\forall x \forall y(xEy \rightarrow \neg x=y)$ in the vocabulary of graphs April 22, 2024Problem: Prove the sentence $\forall x \forall y(xEy \rightarrow \neg x=y)$ in the vocabulary of graphs using the axioms of graph theory. So far: The axioms of graph theory given are antireflexivity and symmetry. Here $xEy$ means there's an edge between the vertices $x$ and $y$. We were told we'd need identity axioms for this […]categoricallystupid
- Proving $\exists x P(x) \rightarrow \forall x P(x)$ from $\forall x\forall y(x=y)$ April 22, 2024Problem: Using identity axioms, prove $\exists x P(x) \rightarrow \forall x P(x)$ from $\forall x\forall y \, x=y$. So far: I'm quite stuck on where to even begin. Working backward, I know we want $P(x)$ and $\exists xP(x)$, but I can't quite see how to write it down. I'm used to proof trees over Fitch-style […]categoricallystupid
- Show that $\exists x \forall y R(x,y) \rightarrow \forall x \exists y R(x,y)$ is not provable by natural deduction April 22, 2024Problem: Show that the formula $$\exists x \forall y R(x,y) \rightarrow \forall x \exists y R(x,y)$$ is not provable by natural deduction. So far: First thought that came to mind was simply considering the natural numbers with an order relation to give a counter-example (considering an assignment such that $s(x)=0$). However now $R(x,x)$ is not […]categoricallystupid
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