# Tag: perfect circle

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- Enumerating all finite models up to isomorphism April 21, 2024For bit of context, please read the answers to this previous question of mine. Given a FO language $L$ with finitely many constant and relation symbols. I don't understand why the set of finite $L$-structures that are isomorphic to each other (I am not sure what it means exactly) is countable, and hence can be […]John Davies
- sets and logic, whether a function is injective and/or surjective? [closed] April 20, 2024Let $f:\mathcal P(\Bbb N)\to\mathcal P(\Bbb N)$ be the function defined by $f(a)=\Bbb N\setminus a$ for every set $a⊆\Bbb N$. Is this function injective? Is it surjective? Justify your answer. How do I go about proving it?confusedmathstudent
- Show that $\vdash (\forall x(C_x\implies M_x)\implies (\exists xC_x\implies \exists xM_x))$ April 20, 2024Show that $\vdash (\forall x(C_x\implies M_x)\implies (\exists xC_x\implies \exists xM_x))$ The steps I've taken: (1) By the deduction rule, it suffices to show: $$\forall x(C_x\implies M_x)\vdash (\exists xC_x\implies \exists xM_x)$$ (2) Again, By the deduction rule, it suffices to show: $$\forall x(C_x\implies M_x), \exists xC_x\vdash \exists xM_x$$ (3) Switching quantifiers: $$\forall x(C_x\implies M_x),(\neg \forall x(\neg […]AmazingAnimal3219
- Reconstructing a closure operator from a set of fixed points April 19, 2024Let $L$ be a lattice, not necessarily complete. We define a closure operator as a function $f\colon L\to L$ which is: idempotent, $f(f(x)) = f(x)$, isotone, $x\leq y \Rightarrow f(x) \leq f(y)$, extensive, $x \leq f(x)$ for each $x,y \in L$. Then the set of fixed points of $f$ is the set of closed elements. […]Jakim
- What inference rule corresponds to the Generalization rule of daily math? April 19, 2024It seems the following inference rules are all correct in the sequent calculus(See Ebbinghaus's book): $\dfrac{\Gamma \phi \dfrac{y}{x}}{\Gamma \forall x \phi}$, if y is not free in $\Gamma \forall x \phi$ $\dfrac{\Gamma \phi}{\Gamma \forall x \phi}$, if x is not free in $\Gamma$ $\dfrac{\Gamma \phi \dfrac{c}{x}}{\Gamma \forall x \phi}$, if c does not occr in […]William
- Propositional Logic Theorem Proof Help: (¬(P↔Q)↔(P↔¬Q)) April 19, 2024Can anyone help me prove this theorem via natural deduction for my class? $$(¬(P↔Q)↔(P↔¬Q))$$ I can solve one side of the biconditional quite easily using conditional derivation, but have no idea how to solve the other way. Below is my proof for one direction of the biconditional. I am not allowed to use any other […]milfordcubicle2
- First order logic, why are the quantifier rules of inference reasonable? April 18, 2024This picture is from the book "Mathematical Logic, 2nd edition, Christopher C. Leary, Lars Kristensen" : I have two questions : Why are the quantifier rules of inference reasonable as they write in the start of the text ? Consider these variations : $$(\{\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: $$\left( \exists \phi : S \rightarrow S' \right) \left(\ \forall a,b \in S \right) \left[ \left( \phi (a⋆b) = \phi(a) […]Karaji
- Is the inference rule another notation for $\implies$? April 18, 2024In mathematical logic, when we write a formula of the metalanguage (such as $A \vdash B$) above an horizontal rule and another formula (such as $\vdash A \to B$), is it the same as writing a meta implication sign $\implies$ between the two formulas?user1313248
- Prove that if a frame is dense then $\Box \Box A \to \Box A $ is valid [duplicate] April 18, 2024I am having trouble with the following question on modal logic. A modal logic frame $M =\langle W, R\rangle$ is dense whenever $$\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

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