Artificial intelligent assistant

a fortiori

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a fortiori
‖ a fortiori, advb. phr. (eɪ fɔəʃɪˈɔəraɪ, fɔətɪ-) [L. a. from, fortiori stronger (sc. argumento).] With stronger reason, still more conclusively.1606 R. Parsons Answer to Coke iii. 52 Yet cannot hee either tacitè or à fortiore..take vnto him, all the power..which the said President and fellowes haue... Oxford English Dictionary
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Argumentum a fortiori
Given some theorem with an additional restriction imposed upon this axiom, an "a fortiori" proof will always hold. Thus, (2) is an "a fortiori" argument. wikipedia.org
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a fortiori
a fortiori/ˌeɪ ˌfɔ:tɪˈɔ:raɪ; ˌeˌfɔrtɪ`ɔˌraɪ/ (Latin 拉) for this stronger reason 因为这一更充分理由; 更加 If he can afford a luxury yacht, then a fortiori he can afford to pay his debts. 他若能买得起豪华的快艇, 就更能匀出钱来还债了. 牛津英汉双解词典
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Fernique's theorem
Then there exists α > 0 such that A fortiori, μ (equivalently, any X-valued random variable G whose law is μ) has moments of all orders: for all k ≥ 0 wikipedia.org
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usuary
usuary Roman Law. (ˈjuːzjuːərɪ) [ad. late L. ūsuāri-us n., f. ūsuārius a., f. L. ūsus use n.] One who has the use but not the ownership of a thing.1871 Poste Gaius iv. 507 As the usufrutuary has no possession, it follows a fortiori that the usuary has no possession. Oxford English Dictionary
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Satake isomorphism
This can be obviously simplified to , which is a fortiori an equivalence of Tannakian categories . Notes References Representation theory wikipedia.org
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Abba Yossi
By way of an inference from minor to major premise (A fortiori), it should not be thought of as weakness for a man to seek assistance in these troubling wikipedia.org
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Topologic Entropy of the the free-square flow S someone knows how to prove that $\eta(n)= \mu^2(n)$ and a-fortiori μ(n), is not deterministic? I'm prove that the system associated to the flow $(X_S,T_S)$ is ergodic, w...
I assume that the notation you are using comes from Sarnak's lecture on Möbius randomness and dynamics. In that case, the result you are looking for appears as Theorem 10 on page 9 of this paper of Sarnak: Three Lectures on the Mobius Function Randomness and Dynamics.
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How to use of the constant " False" in a predicate logic implication appearing in a proof of : X included in EmptySet <=> X is empty While trying to prove the basic fact that : any set included in the EmptySet is empt...
It is the Negation Introduction rule. If the Language has the _falsum_) constant $\bot$, the rule amounts to : > $\text {from } P \to \bot, \text { derive } \lnot P$. In fact, we can define _negation_ from $\bot$ : > $\lnot P \leftrightarrow P \to \bot$.
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Glasser's master theorem
certain integrals over the whole interval from to It is applicable in cases where the integrals must be construed as Cauchy principal values, and a fortiori wikipedia.org
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Samson of Chinon
Ishmael Bet ha-Miḳdash, on the rules for deductions by analogy and conclusions a fortiori Netibot 'Olam, containing explanations of the 32 rules of R wikipedia.org
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Uniform boundedness principle statement Consider the uniform boundedness principle: **UBP**. Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i \in I}$ be a family (not necessarily countable) of continuous linear...
To illustrate the point which Davide explained in his answer, let us have a look at a concrete example. Let us take $E=$ the set of all real sequences with finite support (i.e., only finitely many terms are non-zero). Let us use the norm $\|x\|=\sup_n |x_n|$. The space $E$ is a linear normed space, ...
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Why is a finitely generated $\mathbb Z$-module a finitely generated $\mathcal O_K$-module There is something I don't understand in Neukirch's Algebraic number theorey. He said that: "every ideal is a finitely generat...
If $R \subset S$ are rings and $M$ is an $S$-module which is finitely generated as an $R$-module, then it is finitely generated as an $S$-module too. Indeed, if $m_1, \ldots, m_n$ is a list of generators, then everything in $M$ is a finite $R$-linear combination $\sum r_i m_i$, so it's also a finite...
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Finding a function $f: \mathbb{R} \rightarrow \mathbb{R}$ with this property? > Find a function $f: \mathbb{R} \rightarrow \mathbb{R}$ that is Riemann integrable on every bounded and closed interval, such that the fun...
Just take the function $f(x)=1$ for $x>0$ and $f(x)=0$ for $x\leq 0$. The integral should have a corner at $x=0$.
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Consequence in Logic For arbitrary formulas $A,B,C$ it holds that: 1. $\\{A,B\\} \vDash C $ if $A \vDash (B \Rightarrow C)$ 2. $(A \Rightarrow B) \vDash C$ if $A \vDash (B \Rightarrow C)$ 3. $A \vDash C$ if $A ...
Note that trivially $\vDash p \to p$, so a fortiori $p \vDash p \to p$. But $p \to p \nvDash p$ (suppose $p$ is false). Note that trivially $\vDash q \to q$, so a fortiori $p \vDash q \to q$. But of course $p \nvDash q$.
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