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See : Stephen Cole Kleene, Mathematical logic (1967 - Dover reprint), page 289 : $\cfrac{A, \Gamma \to \Delta, B \quad \quad B, \Gamma \to \Delta, A}{\Gamma \to \Delta, A \equiv B} \equiv \text{: right}$ $\cfrac{A,B, \Gamma \to \Delta \quad \quad \Gamma \to \Delta, A, B}{A \equiv B, \Gamma \to \Delta} \equiv \text{: left}$ If ...

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Question 2 and 4: this is legit. If you prefer to be concrete, first invoke the axiom of infinity (as you say), which states that "there is a successor set", and then use $A_0$ as shorthand for that successor set. (In this proof, it's just assumed that some set exists. To be ultra-formal, you're right: it has to be proved.) You could look at it like this: ...

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If you by "base" set means the left side of the $\vdash$ and you get to use the deduction theorem you can simply use the fact that with contradictions in the premise anything can be proved via reductio ad absurdum (ie axiom 3): $\alpha, \neg\alpha, \neg\neg\beta \vdash \alpha$ $\alpha, \neg\alpha \vdash \neg\neg\beta \rightarrow \alpha$ and similarily ...

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When dealing with collections of random variables indexed by time, we need more than just a probability space $(Ω, \mathcal{F},P)$ to capture the notion of 'information knowable at time $t$'. It is termed a filtered probability space, which is a 4-tuple $(Ω,\mathcal{F},(\mathcal{F_t})_{t\geq 0},P)$ where the new ingredient is a filtration, i.e. a collection ...

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A simple approach is inspired by : George Tourlakis, Lectures in Logic and Set Theory : Volume 2, Set Theory (2003), page 130. The formula $\exists x (x=x)$ can be derived from axiom : $\forall x \varphi(x) \to \varphi[x/t]$ with $\varphi(x) := \lnot x = x$ and $x$ as $t$, by tautological implication [$\vDash_{TAUT} (\mathcal A \to \lnot \mathcal ... 1 Using base set$\{\alpha,\lnot\alpha\}$, from 1. deduce$\beta\to\alpha$and$\beta\to\lnot\alpha$, then apply 3. with$\phi=\lnot\beta$. 1 8 does not divide$x^2-1$is the same as "if 8 divides$x^2-1$, then x is odd" which is the same as: $$8|(x^2-1)\rightarrow 2|(x+1)$$ but $$x^2-1=(x-1)(x+1)$$ Then since$8=4*2$Two divides both$x-1$and$x+1$, and we proven our case 0 Yes, this follows directly from the 2nd incompleteness theorem. By definition, a theory is inconsistent iff it contains both the statement$S$and$\neg S$for some statement$S$. An extension of a theory contains all of the statements of that theory. Therefor if the extension is consistent, then the original theory is as well, since any such statements ... 2 Rewrite $$(x^2-1)=(x-1)(x+1)$$ If 8 does not divide$(x^2-1)=(x-1)(x+1)$, then 2 does not devide$x^2-1=(x-1)(x+1)$. But that is the same as saying that neither$x-1$nor$x+1$are even. Hence$x$must be even. 1 You've tried to prove the converse. The contrapositive is instead "if$x$and$y$have opposite parities, then$x+y$is odd". Let$x$and$y$have opposite parities. Then one is odd and one is even (by, I suppose, something like the pigeonhole principle). Therefore by the division algorithm we may write$x = 2k, y = 2m+1$(wlog that$x$is the even one). ... 0 Let A and B be two points on a circle and let L be the length of the chord AB joining them. Suppose that L is larger than the diameter D of the circle. Consider the diameter through A, and call X the second point where it cuts the circle, so that AX is a diameter, whose length is D. By assumption, L is larger than D, so in particular B and X are different. ... 0 There are, of course, many other ways to proceed. Perhaps a nice way is just to imitate what Euclid's did to prove that there are infinitely many primes, I mean the very first part of the argument, with a slight change: he supposes his set is finite, then multiply all of its members, add 1 to the product, and then proceeds with his proof. Your case is ... 1 The set of positive multiples of$11$can be put in bijection with the set of positive numbers, simply by dividing by$11$. If the former is finite, so is the latter. Can you prove, by contradiction, that there are infinitely many positive numbers? 0 Suppose the set S of all positive multiples of 11 is finite. Let W be the number of digits in base 10 of the largest element in your set S. Now, there is a number with more than W digits that you can construct as follows: N = 1111111111............11111, where you put W+1 or W+2 digits depending on the parity of W: if W is even you take W+2, if it is odd ... 0 Hint: Take a look at Euclid's Proposition IX:21. For "even," think "multiple of$11$." 0 Hint: If the set is infinite, it has a maximum element. If you can find a larger multiple of$11$, you found a contradiction. 5 Saying something cannot be proven by any other means but induction is quite a bold statement, and I am not capable of giving it. It is also not an entirely well defined statement, unless you formally define all your axioms (one of which is induction) and then say "which statements cannot be proven if I remove induction from my set of axioms?" But I ... 1 In set theory the set$\omega$of natural numbers is constructed on base of the axiom of infinity. This as the smallest set that contains$\varnothing$as element and is closed under the operation$a\mapsto a\cup\{a\}$. After that several properties of natural numbers are proved by induction. For instance the fact that all natural numbers are transitive ... 0 Dedekind cuts are somewhat elementary from an analysis point of view, but understanding exactly why they give the (complete) real numbers with all arithmetic operations defined as expected with all the desired properties is a bit harder. For addition, if you have two upper bounded sets of rational numbers then you can form the set of all pairwise sums of ... 3 Consider a similar example from set theory : Kenneth Kunen, The Foundations of Mathematics (2009), page 10 : Axiom 0. Set Existence :$\exists x(x = x)$. Axiom 1. Extensionality :$\forall z(z \in x \leftrightarrow z \in y) \to x=y$. Axiom 3. Comprehension Scheme : For each formula,$\varphi$, without$y$free,$\exists y\forall x(x \in y ...

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I found exactly the same question on Quora and I am just copying the answer given by David Joyce, Professor of Mathematics at Clark University. I hope you will find it useful. Here's the link to the original answer. Axioms come mainly in two different kinds—existential and universal. They often go along with definitions. For instance, an existential ...

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Definitions aren't used to say things exist or something is true about things. They're used to make it easier to talk about things. where as an axiom is a statement or proposition which is regarded as being established, accepted, or self-evidently true.

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A definition is a conservative extension of the language by a new symbol and some axioms involving this symbol. The key word here is conservative; in general axioms strengthen the system in question, while definitions are not allowed to do so.

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From a proof theory perspective, there is no difference. They are both effectively announce the truthhood of something without providing a proof thereof. The difference arises when one applies model theory, which is required to apply the mathematical results (whether applied explicitly or implicitly). A definition is wholly contained within the ...

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A definition is a choice to call something by a specific name/reference/identifier/pointer. What is a name? A cognitive synonym for what people understand to be "equivalent". That is a long philosophical discussion. In any sense and reference, Sinn und Bedeutung for Frege, what part of a name attaches to the thing being named? None. A name is an ablative ...

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Here is my simple explanation of what I think the difference is between a definition and an axiom: An axiom is a rule or a "law of the land" that we decide we will follow/enforce. For example, the axiom of choice in set theory says if you have any arbitrary collection of non-empty sets, you can always form a new set by picking one element from each of the ...

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Axioms are not "defined to be true"; I'm not even sure what that would mean. What they are is evaluated as true. Practically speaking all this means is that in the mathematical context at hand, you're allowed to jot them down at any time as the next line in your proof. Definitions have virtually nothing to do with truth, but are instead shorthand for ...

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An axiom is a special kind of definition. It is a definition of truth in a specific situation. Not all definitions are axioms though. An axiom is a definition that changes what is or is not true.

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A definition is just a name you attach to something. For example you can attach the name "zero" to something. But for the definition to make sense that "something" must exist and that is usually guaranteed by some axiom or theorem.

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In this instance, I'm taking Peano arithmetic to be defined in the first-order theory over functions $0, s, +, \times$ of arity 0, 1, 2, 2 respectively. The symbol $0$ is just that - a symbol. It needs no definition in this language. It already exists. We need the axioms to tell us what we're allowed to do with these symbols. You want to "define" how $0$ ...

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In a sense one can say that are just different forms of postulation, one defines definitions and one defines axioms (see for example equivalence between natural deduction systems and formal axiomatic systems). In a further investigation, there can be (more or less) differences, between the two. For example: Definitions define new concepts (based on ...

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