Tag: “Math Improve”
Logic on Math StackExchange
- proof Integral of All trig function in Riemann Sum form only(try do not use complex number). June 1, 2023want to find a way to prove all trig functions by Riemann Sum form only. Is there a book or paper that has proved it? I am sure there are, but I can not find anything.Issac
- full second order arithmetic: the truth and axiomatizations June 1, 2023What of the natural subsystems of the full second order arithemtic $\mathsf{Z}_2$ is not recursive ? What does it mean for a sentence of the full SOA to be true. I'm interested here in the notion of $\beta_k$ models where we do speak about truth. I would say, it depends, sometimes the set of natural […]user122424
- A syntactic deduction on Monoids' language June 1, 2023Let $\mathcal{L}=\{e,\cdot\}$ be the monoids' language. Give a syntactic deduction about: $$\forall x (x \cdot e = x \land e \cdot x = x) \vdash \forall z (\forall x \hspace{2mm} z \cdot x = x) \rightarrow z=e$$ I am stuck with this, I tried the following: \begin{matrix} \forall x (x \cdot e = x \land […]Superdivinidad
- Why not use the substitution axiom when proving $A=B$ iff $\mathcal{P}(A)=\mathcal{P}(B)$? June 1, 2023In my set theory book we have a proof that $A=B$ if and only if $\mathcal{P}(A)=\mathcal{P}(B)$ where $\mathcal{P}(X)$ is the power set of $X$. The author shows that if $\mathcal{P}(A)=\mathcal{P}(B)$ then $A=B$. And in the case of $A=B$ instead of immediately concluding that $\mathcal{P}(A)=\mathcal{P}(B)$ (from the axiom of substitution) the author proves it as follows: […]emanresu
- How to prove that { ¬, →,$\forall$ } is a functionally complete set of connectors? June 1, 2023I already know how to prove that a set is functionally complete in propositional language without having the quantifier in the set. Now how can I prove tha this set { ¬, →,$\forall$ } is functionally complete in laguage of predicates ?Jazmine
- What are the order types of computable pseudo-ordinals with no c.e. descending chains? May 31, 2023The notion of a “computable pseudo-ordinal”, i.e. a computable linearly ordered set with no hyperarithmetical descending chains, is an old one going back to Stephen Kleene. Joe Harrison wrote the definitive paper on them in 1968, showing that any such linear order is either well-ordered or has order type $\omega_1^{CK}(1+\eta)+\alpha$ for some computable ordinal $\alpha$, […]Keshav Srinivasan
- Investigation on bijection between Axiomatic systems May 31, 2023As an undergraduate math major, I am planning to do a project investigating what conditions are necessary for a bijection to exist between two axiomatic systems. Should I do this topic and with the skills of understanding first order logic, is this an unattainable goal?Nobody It
- Proving $P\rightarrow Q$ is false by assuming $P$ and showing its possible that $\neg Q$? May 31, 2023Im being told my proof is invalid and its driving me crazy because i just dont understand how its not valid. Here: Proving that a theorem is invalid ($\left(A\cap B\right)\triangle C\subset \left(A\triangle C\right)\cap \left(B\triangle C\right)$) In my opinion, the statement is of the form: $P\rightarrow Q$, and what i have done is assumed $P$ and […]lightyourassonfire
- When proving a proposition is false or doing proof by counterexample, do you need to show an explicit example? May 31, 2023To prove a proposition is false, do you need to find an explicit example for which the proposition is not true, or can you just assume stuff and show it leads to a contradiction? For example: Suppose $R_1$ is a total order on $A_1$, $R_2$ is a total order on $A_2$, and $A_1\cap A_2=\emptyset$. Then […]lightyourassonfire
- Is a relation in partional order if its antisymmetric only sometimes? May 31, 2023$A$ = the set of all countries in the world. $R=\\{(x,y)\in A\times A: P(x) \le P(y)\\}$ $P(x)$ means population of $x$ The relation R is antisymmetric only if there are no countries with the same population, so this relation is only antisymmetric sometimes. So I guess the statement "$R$ is a partional order on A$ […]lightyourassonfire
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