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- What inference rule corresponds to the Generalization rule of daily math? April 19, 2024It seems the following inference rules are all correct in the sequent calculus(See Ebbinghaus's book): $\dfrac{\Gamma \phi \dfrac{y}{x}}{\Gamma \forall x \phi}$, if y is not free in $\Gamma \forall x \phi$ $\dfrac{\Gamma \phi}{\Gamma \forall x \phi}$, if x is not free in $\Gamma$ $\dfrac{\Gamma \phi \dfrac{c}{x}}{\Gamma \forall x \phi}$, if c does not occr in […]William
- Propositional Logic Theorem Proof Help: (¬(P↔Q)↔(P↔¬Q)) [duplicate] April 19, 2024Can anyone help me prove this theorem via natural deduction for my class? $$(¬(P↔Q)↔(P↔¬Q))$$ I can solve one side of the biconditional quite easily using conditional derivation, but have no idea how to solve the other way. Below is my proof for one direction of the biconditional. I am not allowed to use any other […]milfordcubicle2
- An example with moving a premise of an inference rule into the context April 19, 2024In a way, this is a continuation about this question, it also partially motivates this question. Consider the following universes:$$\frac{\Gamma\vdash i:I}{\Gamma \vdash A_i\text{ type}}\quad \frac{}{\Gamma\vdash B=^{\text{def}} \Sigma(i:I)A_i\text{ type}}\\ \frac{\Gamma \vdash i:I\quad \Gamma \vdash a:A_i}{\Gamma\vdash T_i(a)\text{ type}}\quad \frac{\Gamma\vdash b:B}{\Gamma\vdash T(b)=T_{\pi_1(a)}(\pi_2(b)) \text{ type}}$$ Now suppose I want to get rid of $\Gamma\vdash i:I$ in the premises. Instead, […]user837242
- First order logic, why are the quantifier rules of inference reasonable? April 18, 2024This picture is from the book "Mathematical Logic, 2nd edition, Christopher C. Leary, Lars Kristensen" : I have two questions : Why are the quantifier rules of inference reasonable as they write in the start of the text ? Consider these variations : $$(\{\psi\rightarrow \phi\}, \psi \rightarrow (\exists x \phi)\})$$ $$(\{\phi\rightarrow \psi\}, (\forall x \phi) […]user394334
- Logical Definition of Isomorphism April 18, 2024I'm trying to define isomorphism in a logical way. Is the following statement true for Isomorphism's definition? Let $\langle S,⋆\rangle$ and $\langle S',⋆'\rangle$ be Algebraic Structures. These two structures are isomorphic if and only if: $$\left( \exists \phi : S \rightarrow S' \right) \left(\ \forall a,b \in S \right) \left[ \left( \phi (a⋆b) = \phi(a) […]Karaji
- Proof by contradiction square root of 3 is irrational [closed] April 18, 2024I think it requires circular thinking Please see this YouTube video https://www.youtube.com/watch?v=XQn47l2bt1g for details I found this alternative approach promising https://math.stackexchange.com/a/930492/1224828, and would like to know what is wrong with the approach in the YouTube video please.Caleb
- Is the inference rule another notation for $\implies$? April 18, 2024In mathematical logic, when we write a formula of the metalanguage (such as $A \vdash B$) above an horizontal rule and another formula (such as $\vdash A \to B$), is it the same as writing a meta implication sign $\implies$ between the two formulas?Anserin
- Prove that if a frame is dense then $\Box \Box A \to \Box A $ is valid [duplicate] April 18, 2024I am having trouble with the following question on modal logic. A modal logic frame $M =\langle W, R\rangle$ is dense whenever $$\forall x\forall y(Rxy\rightarrow\exists z(Rxz \land Rzy))$$ Prove that the formula $$\Box\Box A\to\Box A $$ is valid in every dense frame. We don't make any assumptions about the frames. How can this be done?A_alk
- Confusion on exercise on $(\mathbb Z, +)$ in Buechler's Stability theory book April 18, 2024Given as exercise 2.5.1: Show that the complete type realized by $1$ in $(\mathbb Z,+)$ is non-isolated. HINT Use the preceding exercise. The previous exercise: Let $\vec a$ and $\vec b $ be finite sequences from $\mathcal M.$ Prove that $\operatorname{tp}_{\mathcal M}(\vec a\vec b)$ is isolated iff $\operatorname{tp}_{\mathcal M}(\vec a/\vec b)$ and $\operatorname{tp}_{\mathcal M}(\vec b)$ […]spaceisdarkgreen
- Are there any problems about the difference between set theoretic definitions of polynomials? April 17, 2024I am a novice about this question, so if there is a misunderstanding then I apologize for it. As for Peano axioms, if I choose Zermelo natural numbers, and you choose von Neumann ones, then this doesn't cause any problems because we can prove Peano axioms are categorical i.e. it is guaranteed my model, and […]categoricalequivalent
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