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1

The inference is licensed by the Natural deduction rule of Existential elimination: ($\exists E$) If $Γ⊢∃zφ$ and $φ[y/z]⊢ψ$, then $Γ⊢ψ$, provided that $y$ does not occur in $Γ,φ,ψ$. The argument runs as follows: we know that there is $\ln x$ [i.e. $\exists z \ (z=\ln x)$] and we use the assumption that $y=\ln x$ [this is: $φ[y/z]$] , where $y$ is a new ...


2

Hint: The base case is easy, so let's look at the inductive step. Assuming $P(m)$ is true, let's consider any set $S$ of $2m+3$ distinct integers from among $[-2m-1, 2m+1]$. If less than $3$ of those are from among $R = \{\pm(2m+1), \pm2m\}$, the induction hypothesis takes care of it. So that leaves cases where $|S \cap R| \in \{3, 4\}$. Here use ...


2

Given your level, I would say that an excellent book would be "Linear Algebra Done Right" by Axler. His book is less computational than most and a great second proof-based course in LA. ...


1

Without a doubt, "Linear Algebra" by Serge Lang. (A note of caution-there is also "'Introduction to Linear Algebra" by Lang-a fine book by itself, but from your question, you are looking for a more advanced text, like the first.)


4

The book says "We now explain why the Riemann hypothesis is plausible on probabilistic grounds". That answers the question. We cannot assume that RH is true "due to high probability of being true". It is just one more indication that we should believe in it. A proof, however, is something different.


2

When $\delta(\varepsilon)$ is written as you have above, it is merely a notational reminder that our choice of $\delta$ has to depend on the $\varepsilon$ we're given -- nothing more. In fact, $\delta$ also depends on $f$, $a$, and $L$. Writing $\delta(\varepsilon)$ does not mean that $\delta$ is a function to which we may plug in $\varepsilon$ to get our ...


3

The reason why it's enough to find a single $\delta$ is: If you've found a $\delta$ that meets the requirement, then every positive $\delta'$ that is less than $\delta$ also meets the requirement. Reason: If $|x-a|<\delta'$ and $\delta'<\delta$, then for sure $|x-a|<\delta$, so continue on to the conclusion regarding $f(x)$ vs $L$ that was justified ...


0

Proof 2 is certainly a valid proof, but is not very enlightening. Where did that value for $\delta$ come from? Proof 1 is more likely to convince your professor that you have done your work. The first part of the proof, which is fine, says essentially that if $|x-a|<\delta$ then $|f(x)-L|< g(\delta)$, where $g$ is some function of $\delta$. Now, in ...


1

"..from what I can see, proof methods #1, and #2 are essentially the same..". Yes,basically they are. But concerning the fundamental question of rigourousness, well, what is more rigorous than using the definition to prove the desired result as the $\epsilon - \delta$ approach does? You may argue-and my feeling is that your "healthy confusion" stems from ...


1

Bernard's answer highlights the key algebraic step, but I thought I might mention something that I have found useful when dealing with induction problems: whenever you have an induction problem like this that involves a sum, rewrite the sum using $\Sigma$-notation. It makes everything more concise and easier to manipulate: \begin{align} ...


2

You're supposed to prove that $$(k+1)!-1+(k+1)(k+1)!=(k+2)!\,\color{red}{\mathbf -}\,1$$ Simplifying both sides by $(k+1)!$ yields $$1+(k+1)=k+2,$$ which is pretty obvious, isn't it?


1

If I get you right, then what you refer to as global state is usually just called an axiom system. Such an axiom system is simply a set of formulas, which are assumed to hold without any further proof needed. Now if we have a sequent $$ \Gamma_0 , \Gamma \to \Delta $$ where $\Gamma_0$ is a sequence of formulas from an axiom system $G$, then we say that $$ ...


5

If $Z$ is the integers of a nonstandard model of PA, let $n\in Z$ be any nonstandard positive element and consider $2^n$. Then the ideal $(2^n,2^{n-1},2^{n-2},\dots)\subset Z$ generated by $2^{n-m}$ for all standard natural numbers $m$ is not finitely generated. More generally, any element of $Z$ with a nonstandard number of prime factors gives rise to a ...



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