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- Real-life example of fail by not using equivalent transformation December 7, 2024Solve the equation : $\sqrt{x}=x-2$ Square it : $x=(x-2)^2$ Expand : $x=x^2 - 4x +4$ Everything to the left side : $x - (x^2 - 4x +4) = 0$ Refine the equation : $x^2 - 5x +4 = 0$ Factorize : $(x-1)(x-4) = 0$ Getting roots : $x =1$ or $4$ There is nothing wrong […]imida k
- Do propositional calculus identities have an implied $\forall$ when used in first-order logic contexts? December 7, 2024This is a list of identities in propositional calculus. However, if you were to apply these in first-order logic, would you have to put a $\forall$ for each relevant entity ($P$’s, $Q$’s, etc.)? If so, is it still possible to use those identities to define the logical elements themselves? If not, why not?Mathemagician314
- How do you construct a While loop using propositional dynamic logic? December 7, 2024I am learning some modal logic from a textbook (Modal Logic from Blackburn, De Rijke and Venema) and they have an example about propositional dynamic logic (PDL). Here they explain that if $\pi$ is a program, then $\pi^*$ is also a program, and $\pi^*$ is an iteration of $\pi$ for a finite (possibly zero) number […]UnrulyTank
- Proof of Deduction Theorem in Hilbert-Calculus, Beginner friendly December 7, 2024Deduction Theorem: Let $ \Sigma$ be an axiom system, and let $ \psi, \varphi$ be $ L$-formulas. Then the following always holds: $ \Sigma, \psi \vdash \varphi \iff \Sigma \vdash \psi \rightarrow \varphi $ “$ \Rightarrow$” We prove by strong induction on the length of a proof $ n \geq 1$ that for arbitrary $ […]God
- Proof of the General Deduction Theorem December 6, 2024Theorem: For all axiom systems $\Sigma$ and all formulas $\varphi, \psi_1, \ldots, \psi_n$, it holds that: $ \Sigma, \psi_1, \ldots, \psi_n \vdash \varphi \iff \Sigma \vdash \psi_1 \to \ldots \to \psi_n \to \varphi$ Proof: $\Rightarrow$: We prove by complete induction. Initially, that is, for $n = 1$, it holds: Let $\Sigma, \psi_1 \vdash \varphi$, and […]God
- Existence of a model of $\mathsf{ZFC}$ which has the internal standard model of arithmetic coincide with the real standard model of arithmetic December 6, 2024$\mathsf{ZFC} + \operatorname{Con}(\mathsf{ZFC})$ can prove that there is a model $M$ of $\mathsf{ZFC}$ which has the natural numbers structure $\mathbb{N}^M$ as the standard model of arithmetic inside $M$ but $\mathbb{N}^M \ncong \mathbb{N}$ holds outside, where $\mathbb{N}$ means the real standard model of arithmetic in the meta-theory $\mathsf{ZFC} + \operatorname{Con}(\mathsf{ZFC})$. Then, can we take a model […]Pointwise
- Do "truly" infinite proofs exist? December 6, 2024An assumption underlying this earlier question was the existence (and greater expressive strength) of infinite proofs in logics like $\mathcal{L}_{\omega_{1}^{CK}, \omega}$ (based on, for example, the discussion in §2 chap. IX of Mathematical Logic by Ebbinghaus et. al., and answers like this and this) However the comments in response to the question have led me […]NikS
- How do you define universal and existential quantification rigorously and non-circularly? December 5, 2024Universal and existential quantification are necessary to build math from set theory. But I can't think of any way to define them besides the traditional intuition: that is, $(\forall x\in S)[p(x)]$ means "for all $x$ in $S, p(x)$" and $(\exists x\in S)[p(x)]$ means "there is some $x$ in $S$ such that $p(x)$". I know that […]Mathemagician314
- Do quantified statements without domains mean anything? What is one to do when it seems impossible for there to be any domain? [duplicate] December 5, 2024I used to find quantified statements without domains perfectly sensible, but now I can hardly grasp them. Such statements are in introductory texts, proof assistants, and many other places. When a statement concerns numbers, it almost always has a domain, and there is no ambiguity. In many other statements, however, there is less clarity. How […]interested
- Define if the statement: ∀x(A(x) ≡ B(x)) ~ (∀x A(x) ≡ ∀x B(x)) is always true December 5, 2024There is a statement: $$∀x(A(x) \equiv B(x)) \sim (∀x A(x) \equiv ∀x B(x))$$ that seems to be true in all models, but unfortunately I can't think of an idea to proove that, except this, but I am not sure whether it makes sense... $$∀x(A(x) \equiv B(x)) \sim (∀x A(x) \equiv ∀x B(x))$$ is the same […]Artkol
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