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- Suppose $R$ is a relation on $A$. Let $S$ be the transitive closure of $R$. Then $Dom(S)\subset Dom(R)$ June 8, 2023$Proof$. Suppose $x\in Dom(S)$ and $x\notin Dom(R)$. Then there is some $y\in A$ such that $xSy$. Since $S$ is the transitive closure of $R$, there exist some set of pairs in $R$ such that $x$ and $y$ can be connected. But $x\notin Dom(R)$, so this connection cannot be made. Therefore it must be that $x\in […]lightyourassonfire
- Is Tarski's exponential function problem arithmetically decidable? June 8, 2023https://en.wikipedia.org/wiki/Tarski%27s_exponential_function_problem shows a very interesting problem, as for me begging for an undecidability proof (as the Tarski-Seidenberg theorem itself is already very strong, and this would strengthen it a lot more) But is it decidable with an oracle for arithmetical truth? (Given an arithmetical sentence, this oracle will return whether it is true or false) […]Alexey Slizkov
- Is there some variant of sequent calculus that allows for non-trivial axioms? June 8, 2023Apologies if the question doesn't make sense: it's one of those cases where my confusion is so diffuse that I'm not even sure how to ask the question. In short, I would like to know if it's possible, within some variant of sequent calculus, to prove the sequent $$ \varGamma \,\vdash\, V $$ starting from […]pglpm
- When does a proof have too little details? (prove that the reflexive closure of a strict partial order is a partial order) June 8, 2023Suppose $R$ is a strict partial order on $A$. Let $S$ be the reflexive closure of $R$. Then $S$ is a partial order on $A$. $Proof.$ $R$ is transitive and antisymmetric, and $i_A$ is reflexive. So since $S$ is just $R\cup i_A$, $S$ is a partial order on $A$. $\square$ Does this proof have too […]lightyourassonfire
- Henkin's Model Existence Theorem and Forcing June 8, 2023Henkin's Model Existence Theorem says that, if $T$ is a consistent theory, then there exists some set $S$ such that $S \models T$. Suppose we are trying to prove the theorem $\text{Con}(ZFC) \rightarrow \text{Con}(ZFC + \neg CH)$. Are the following steps a correct way to start? Assume $\text{Con}(ZFC)$. Use Henkin's Model Existence Theorem to get […]Gavin Dooley
- Doing a commentary of a proof (if a poset has a unique minimal and that minimal is not the smallest, then the poset is infinite) June 8, 2023The initial proposition was: "If a subset of a partially ordered set has exactly one minimal element, must that element be a smallest element?" And here is a proof: https://math.stackexchange.com/a/2280977/1184854 But this doesnt directly proof the proposition right? They prove a little bit different proposition. Here is my commentary of the proof: The proposition is […]lightyourassonfire
- Expressability in notation of set theory June 8, 2023The following from Terence Tao's blog dated 2007-08-27: It seems that one cannot express For every x and x’, there exists a y depending only on x and a y’ depending only on x’ such that Q(x,x’,y,y’) is true (*) in first order logic! https://terrytao.wordpress.com/2007/08/27/printer-friendly-css-and-nonfirstorderizability/#more-172 Does the following adequately formalize this statement using the notation […]Dan Christensen
- Using Logic to prove the Infinity of Primes June 7, 2023Can you prove the infinitude of primes using logic? I ask because of the following. Let P signify the set of prime numbers. Let A be a subset of P. $\forall p\left(p\in P\right)$. Let F(A) signify that A is finite. $\exists A\exists p (F(A)\rightarrow p\notin A)$. This statement is true because I can construct an […]AutistandProud
- Is $x<y$ only meaningful if $x$ and $y$ are elements of the same set? June 7, 2023$xlightyourassonfire
- prove -(i=0) => suc(prd(i)) = i June 7, 2023Using peano second axiom $∀i, (i = 0) ∨ (∃j, i = suc(j))$ how to prove $∀i : Nat, ¬(i = 0) ⇒ suc(prd(i)) = i$ ? given $spec \space \text{Nat_with_prd} = Nat \space then \\ op\space prd : Nat →? Nat \\ ∀i : Nat \\ • def \space prd(i) ⇔ ¬(i = 0) […]Kim
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