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The standard axioms for identity (or equality) are : I.1 $∀x \ (x = x)$ I.2 $∀x∀y \ (x = y → y = x)$ I.3 $∀x∀y∀z \ (x = y ∧ y = z → x = z)$ I.4 $∀x_1 \ldots x_n y_1 \ldots y_n \ (x_1 = y_1 \land \ldots \land x_n = y_n → t(x_1, \ldots ,x_n) = t(y_1,\ldots, y_n))$, $\ \ \ \ \ ∀x_1 \ldots x_n y_1 \ldots y_n \ (x_1 = y_1 \land \ldots ... 0 The paradox arose from an early attempt to axiomatize set theory. Using this system, it was possible to both prove and disprove the existence of the Russell Set. You could prove its existence by using an axiom of unrestricted comprehension. You could disprove it by using only the rules of FOL. So, simply eliminating unrestricted comprehension alone would ... 0 I'd say that not only axiom schema of specification helps in fighting with paradoxes, but the whole axiomatization. The existence of the set in Russel's paradox$R = \{x|x\not\in x\}$cannot be proven via axioms of ZFC, roughly because it uses specification and the latter is restricted to subsets of other sets. Furthermore, using axiom of regularity, we ... 0 The point with the formal axioms of set theory is that whenever we want to create a new set, we need to start from an old set. Russels paradox uses the "set" $$S=\{x: x\notin x\}$$ However$S$does not take its elements from an already existing set, which is required by the Axiom of separation. Thus this does not need to be a set, as it was not constructed ... 2 See Russell's Paradox : Zermelo replaces NC [Naïve Comprehension principle] with the following axiom schema of Separation (or Aussonderungsaxiom): $$∀A ∃B ∀x (x \in B \iff (x \in A \land \varphi)).$$ Again, to avoid circularity,$B$cannot be free in$\varphi$. This demands that in order to gain entry into$B$,$x$must be a member of an ... 3 Borrowing from Kevin's answer, but working around the second point that is not motivated:$\phi\rightarrow(\neg\neg\phi\rightarrow\phi)\tag{axiom 1}\neg\neg\neg\phi\rightarrow(\neg\neg\phi\rightarrow\phi)\tag{axiom 10}$Now to use axiom 8 we will need the$\phi\lor\neg\neg\neg\phi$instead of axiom 11. To do that we will need ... 0 Apply 8. with$\varphi = \varphi$,$\psi = \lnot\varphi$and$\chi = \lnot\lnot\varphi \to \varphi$. By modus ponens, you have to show three things$\varphi \to (\lnot\lnot\varphi \to \varphi)$: that is axiom 1.$\lnot\varphi \to (\lnot\lnot\varphi \to \varphi)$: that is a kind of 10, where the negation is on second term instead of first. ... 0 The mistake here is to say that$\mathcal S=\left\{ w : |\{n : S_n(w) > \sqrt{2 n \log \log n}\}| = \infty \right\}$is at most countable. As a counterexample, consider the set$\mathcal S'$of all real numbers in$(0\,,1)$whose binary expansion is$0.0^{n(0)}1^{n(1)}0^{n(2)}1^{n(3)}0^{n(4)}...$, in which a string of$n(0)0$s is followed by$n(1)$... -2 I am inclined to believe that the answer to my own question: "Concept of “eventually almost surely” as an artefact of measure-theoretic axioms?" is yes. And I will try to work through an explicit example to demonstrate it. Start with Law of the iterated logarithm and coin tossing. An example outcome of an experiment is just a "string" ... 2 No, you can easily prove that there are infinitely many prime numbers without appealing to the axiom of infinity. The usual proof by Euclid does just that. The axiom of infinity is used to show that the collection of natural numbers is a set, and therefore the collection of prime numbers is a set as well. 0 While some texts use 'disjoint' to mean 'mutually disjoint', these texts seem to use 'disjoint' as meant to be 'pairwise disjoint'. 1 I just wanted to add one thing to BrianO's nice answer. It's certainly strange to say that the existence of$z\cap x$should be proved by Replacement, when it most naturally follows from Separation, as in BrianO's answer. It's strange, that is, unless Separation doesn't appear on your list of axioms. I don't have a copy of the book by Prestel and Delzell, ... 3 In fact you (and they) don't need to use$\cap$in order to state Regularity. As they haven't defined the symbol, they shouldn't use it. You can eliminate intersection simply by expanding its definition: $$\text{AxRegularity}\iff \forall x (x\neq\emptyset \longrightarrow \exists z (z\in x \wedge \forall u \neg (u\in z \wedge u\in x))$$ Their statement ... 2 We use definition from the question: pure truths (provable and not disprovable), to answer the question Are there "Peano complete" paraconsistent arithmetics, where every theorem of Peano arithmetic is a pure truth? For every formula$\phi$, the longer formula$E_\phi \equiv \lnot (\phi \land \lnot \phi)$is a theorem of PA. In any ... 1 Here is a derivation:$(k\to (k\to k)) \to (\neg(k\to k) \to \neg k)$by Ax.2, with$X = k, Y = (k\to k)$.$k\to(k\to k)$by Ax.1, with$X=k, Y=k$.$(\neg(k\to k) \to \neg k)$by 1. and 2. via MP.$(\neg(k\to k) \to \neg k) \to [((\neg k\to k)\to k) \to (\neg(k\to k) \to \neg k)]$Ax.1 with$X = (\neg(k\to k) \to \neg k) =$3.,$Y = ((\neg k\to k)\to ...

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Contingency can be somehow defined in terms of necessity: $$\operatorname{C}p\leftrightarrow(\lnot\operatorname{N}p\,\land\,\lnot\operatorname{N}\lnot p)$$ You can add the above (defining) axiom to a system which you use to formalize necessity and have system for both necessity and contingency. Note that the "meaning" of the above axiom is just "Contingency ...

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Yes, one can define an affine space over a ground field $\Bbb F$ to be a nonempty set $\Bbb A$ endowed with maps $$\mu: \Bbb A \times \Bbb A \times \Bbb A \to \Bbb A$$ and $$\Lambda: \Bbb F \times \Bbb A \times \Bbb A \to \Bbb A$$ that together satisfy a particular list of reasonable axioms. Informally, we should think of these maps as $$(x, y, z) \mapsto x ... 1 Put z = \frac{x+y}{2} (the field can't have characteristic 2 except in the trivial case when x > y is always false). Then x > z > y. 1 To my knowledge, the first postulate is that given any two points, there is a line which has them as endpoints; this has little to do with curves, so how could the definition violate that? Actually, a definition in itself cannot violate any statement at all. Even Definition. A pair of two points is called a Jabberwocky if there does not exist a line ... 0 There is no need to introduce angle measure. We use only the undefined intermediacy relation of point triplets. Also, we need the axioms describing the undefined concept of congruence making it possible to copy triangles (angles). Finally, we need the concept of perpendicularity. See the following figure. The definition of the relation \beta>\alpha ... 0 A concrete example of the process of axiomatization... For millennia, humans have been counting things, recording their number, adding and subtracting them, etc. They were even able prove quite sophisticated theorems establishing various properties of numbers without the use of any formally stated axioms. To introduce more rigour into the proofs of these ... 5 Some useful comments from : Ethan Bloch, Proofs and Fundamentals : A First Course in Abstract Mathematics (2nd ed - 2011); [page 47] Today, virtually all branches of pure mathematics are based on axiomatic systems, and work in pure mathematics involves the construction of rigorous proofs for new theorems. Much of the great mathematics of the past has ... 1 I don't understand what you're doing from the point you introduce 1/a. You've shown that ¬(a≤0∧b≤0) is equivalent to a>0∨b>0. Similarly ¬(0≤a∧0≤b) is equivalent to a<0∨b<0. So we're left with ¬(0≤a∧0≤b) and ¬(a≤0∧b≤0) is equivalent to: (a<0∨b<0) and (a>0∨b>0). From here just show that ab<0, and you've ... 2 The enlargement axiom does not trivialise the theory. It does imply that X is a neighbourhood of every x \in X: every x \in X has some N \in N(x) (this is actually part of the axioms/conditions, often forgotten: every N(x) is non-empty) and N \subseteq X so X \in N(x) by axiom 2. And everything inbetween N and X is a neighbourhood too. We ... 0 The proof is standard: consider c=(1+1)(a+b). Then, by distributivity on the right,$$ c=(1+1)a+(1+1)b $$By distributivity on the left,$$ c=(1a+1a)+(1b+1b) $$By the property of 1 and associativity,$$ c=a+(a+(b+b))\tag{1} $$On the other hand, by distributivity on the left,$$ c=1(a+b)+1(a+b) $$and, by the property of 1 and associativity,$$ ...

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With you idea, you can prove it directly, ie starting with $a+b$ and ending with $b+a$. $$a+b$$ Add $0$ to this equation $$a+b=a+b+(-(b+a))+(b+a)$$ Suppose you previously proved that $-(b+a)=(-b)+(-a)$ $$a+b = a + b +(-b) + (-a) + (b+a)$$ $b + (-b) = 0$ $$a+b = a + (-a) + (b+a)$$ Same for $a + (-a) = 0$ $$a+b = b+a$$

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Yes, it is correct, though a little approximate. $a+b=b+a\\ \equiv\\ a+b+(-1)(b+a)+(b+a)=(b+a)+(-1)(b+a)+(b+a)\\ \equiv\\ a+b-b-a+b+a=(b+a)(1-1)+(b+a)\\ \equiv\\ a-a+b+a=b+a\\ \equiv\\ b+a=b+a$. You must subtract and add the term $b+a$ (i.e. add $0$), otherwise you need to invoke the rule $a=b\equiv a+c=b+c$, which isn't taken for granted. For brevity, ...

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Yes, this is a valid proof, if you state tha each pair of consecutive equations is equivalent. _Then the first statement is equivalrnt to the last one, and the last one is apperently true. But it is simpler to read if you begin with the last ( or better the next to last) statement and write a sequence of statents where a statement is implied by its ...

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The real reason is to avoid degenerate cases. Sometimes the properties are given as: Each two distinct points of P are incident with a unique line. Each line is incident with at least three points. Each two distinct lines of P meet in a unique point. There are four points of P with no three of them collinear. In other words, we want our ...

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