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This is really a question of language and logic. The set of peaks for a given sequence may be finite (including empty) or it may not be. Either the statement it is finite is true or it is false. If it is false then you are in case 2: there are infinitely many peaks. If it is true then you are in case 1.


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Regarding I), the "potential supply" of variables must be "unlimited", basically for two reason : in order to be able to build formulae satisfiable in domain with any finite number of elements; in order to avoid problems with substitution; if an attempted substitution can cause problems due to the "clash" with existing quantifiers, an unlimited supply of ...


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Based on the article "Beweistheoretische Erfassung der unendlichen Induktion in der Zahlentheorie" von Kurt Schütte (Mathematische Annalen 122, 1951, pp. 369-389), the proof-theoretic ordinal for PA+TI($\epsilon_0$) should be $\epsilon_1$.


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Take $\epsilon=1/2$ for example, suppose there is an $N$ such that $\forall n\forall m:n,m\ge N\implies |a_n-a_m|\le 1/2$. Pick up $n$ odd and so big that $n\ge 20$ and $n\ge N$. We then have $a_n-a_{n+1}=1-1/n-4/(1+n)>1/2$ and we found our desired contradiction.


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It seems to me that you have to start (top-down) from : $(P⊃Q),(Q⊃⊥),P \Longrightarrow ⊥$ and then apply $R \supset$ to get : $(P⊃Q),(Q⊃⊥) \Longrightarrow (P \supset ⊥)$ followed by $L \land$ to get : $((P⊃Q) \land(Q⊃⊥)) \Longrightarrow (P \supset ⊥)$. The "final step" must be $R \supset$ again, to reach : $\Longrightarrow ...


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See : Sara Negri & Jan von Plato, Structural Proof Theory (2001), page 49; in the sequent calculus G3cp of classical propositional logic, the logical axioms are : $P, \ \Gamma \Longrightarrow \Delta, \ P$ where $\Gamma$ and $\Delta$ are finite multisets (they can be empty). Thus, $A, B, C \Longrightarrow C, A$ is an axiom, where ...


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$$\int_{a_i}^G\left(\frac1t-\frac1G\right)dt=\int_G^{a_i}\left(\frac1G-\frac1t\right)dt\;,$$ so the first inequality is equivalent to $$\sum_{i=1}^kp_i\int_G^{a_i}\left(\frac1G-\frac1t\right)dt+\sum_{i=k+1}^np_i\int_G^{a_i}\left(\frac1G-\frac1t\right)dt\ge 0\;.\tag{1}$$ The summations on the lefthand side of $(1)$ have identical summands, so they can be ...


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I'm gonna simplify the problem by assuming that $\ell=x_0=0$: notice that $$ \dfrac{f(x_0+t)+f(x_0-t)}{2}-\ell = \dfrac{(f(x_0+t)-\ell)+(f(x_0-t)-\ell)}{2}. $$ Hence Dini hypothesis is the same as saying that the function $f_1(t) = f(x_0+t)-\ell$ satisfies $$ \int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right|\dfrac{dt}{t}<\infty. $$ As it's proved later ...


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I think this article might help you. Pointwise Convergence of Fourier Series, Charles Fefferman, Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 551-571: http://www.jstor.org/discover/10.2307/1970917?sid=21105651264483&uid=4&uid=2&uid=3737760


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This may give a more satisfactory answer to the OP's question, albeit the problem statement is mildly different. Problem: If $z_1,\ldots,z_n$ are complex, prove that $$ |z_1+\cdots+z_n| \leq |z_1|+\cdots+|z_n|. $$ Solution. We proceed by induction on $n$. For $n=1$, there is nothing to show and the case $n=2$ is the triangle inequality which is given by ...


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Take a look at the proof by Rudin (if $x, y \in \mathbb R^2$, then $|x+y| \le |x|+|y|$): $$|x+y|^2 = (x+y)(x+y) = x \cdot x + 2 x \cdot y + y \cdot y \le |x|^2 + 2|x||y| + |y|^2 = (|x|+|y|)^2$$ The inequality $|xy| \le |x| |y|$ is a consequence of the Cauchy-Schwarz inequality.


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No, and there is no slowest growing function which cannot be proven total by PA. Let $f$ be any recursive function that cannot be proven total by PA; we will require that $f$ be strictly increasing to avoid some difficulties. Then choose a fast growing function $g$ which can be proven total by PA (faster than $g(n) = n$, of course). Let $h(n) = \sup ...


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Using your definition of "grows faster" in the comments, for any primitive recursive increasing function $f(n)$, $f(2n)$ grows faster. (or $f(n^2)$, $f(2^n)$, etc.) If you want much faster, take any primitive recursive function $f$; it will be slower growing than $g(n) = 2 \uparrow^m n$ for some $m$. So increase $m$ greatly; for example, $h(n) = 2 ...


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From $p^3+pq^2+q^3=0$, discuss the four possible cases: $p$ odd, $q$ odd. In this case the three terms are all odd. They cannot sum up to $0$. $p$ odd, $q$ even. In this case, we have one odd term and two even terms. The sum must be odd, so they can not add up to $0$. I guess you can now continue with the other cases now and conclude that they have ...


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This is implicitly a universal statement. To disprove a universal statement, all you have to do is provide a counterexample. So, all it takes to disprove this statement is to find two irrational numbers that their product is rational. Can you find such a counterexample?


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While in conventional language, a contradiction is a statement of the form "X and not X", for mathematical purposes, any false statement is equal to a contradiction. In logic "summer is cold" equals "today is Tuesday and today is not Tuesday" since both equal "false". You don't actually have to form a conclusion of the form $\lnot R \land R$, just ...


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If $f(n)$ is primitively recursive, then so is $f(n)+1$, which, under any reasonable definition I can think of, grows faster than $f(n)$.


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When you say you are working in some (unspecified) logical framework, you are not really telling us what logic you are actually working in. However, in type theories that allow empty types, your rule is not admissible: declaring $y$ to be a member of an empty type is a nontrivial assumption. If you logic does have all types non-empty, then your rule is ...


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There is such a thing as infinitary logics, where a proof can consist of infinitely many steps, suitably arranged. But that is not really mainstream, in the sense that it is not intended to be a useful model of the ordinary mathematical reasoning that is done outside mathematical logic. More mainstream, you may have heard of Gödel's incompleteness theorem, ...


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By the usual definition, a proof is a finite sequence of statements in some language, where each statement is of finite length. However, there is such a thing as Infinitary logic.



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