I think this is a definition of = because it explains the meaning of the new symbol =. However, I wonder why the set theory thinks this as an axiom rather than a definition.
The axiom of extensionality is:  ${\forall}a({\forall}b({\forall}c(((c{\in}a){\iff}(c{\in}b)){\iff}(a=b))))$  The definition of $(P{\iff}Q)$ is:  ${\lnot}((P{\implies}Q){\implies}({\lnot}(Q{\implies}P)))$  or, in other words  $(P{\land}Q){\lor}({\lnot}P{\land}{\lnot}Q)$  In the case of the statement $((c{\in}a){\iff}(c{\in}b))$, this leads us to two options: $((c{\in}a){\land}(c{\in}b))$ and $({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b))$. So far nothing weird. Now, let's see all the possible results of the axiom.  1.$((c{\in}a){\land}(c{\in}b)){\land}(a=b)$ 2.${\lnot}((c{\in}a){\land}(c{\in}b)){\land}{\lnot}(a=b)$ 3.$({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b)){\land}(a=b)$ 4.${\lnot}({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b)){\land}{\lnot}(a=b)$  Coincidentally, they are also all possible statements regarding the statements $((c{\in}a){\land}(c{\in}b))$ and $(a=b)$: both parts are true, the first part is true and the other is false, the second part is true and the first is false and neither is true. The axiom seems not to state anything new, because rephrased it's just produces a list of all possible options regarding the two parts of the statement. Worse, it doesn't seem to say anything about equality, since it allows for $a=b$ with both $(c{\in}a){\land}(c{\in}b)$ and ${\lnot}((c{\in}a){\land}(c{\in}b))$. So, what is the point of this axiom? And in what way does it say anything about equality?
Explain proof by contradiction. I am trying to think/know about something, but I don't know if my base premise is plausible. Here we go.  Sometimes when I'm talking with people about pure mathematics, they usually dismiss it because it has no practical utility, but I guess that according to the history of mathematics, the math that is useful today was once pure mathematics (I'm not so sure but I guess that when the calculus was invented, it hadn't a practical application).  Also, I guess that the development of pure mathematics is important because it allows us to think about non-intuitive objects before encountering some phenomena that is similar to these mathematical non-intuitive objects, with this in mind can you provide me historical examples of pure mathematics becoming "useful"? So, I'm working on a control station that has buttons. It is a one-mesh creation, that means, I don't want to create buttons and things like that separated from the main mesh, but as you can see, I have problems. I would like to make the highlighted parts to be one with the main part around them. I've tried everything, and I have this problem on multiple creations. Any idea?  Update: I've added another example of the problem. I want them to be one face, instead of two. (2nd picture) I'm heading to India next week and have a connection in Munich, Germany. I made the mistake of booking a flight with only an hour and a half layover (freaking out!) and I'm wondering if I'll have to go through customs in Munich or if I can pass right on through. Remotely changing hard drive boot order, windows 7 I have one workstation chassis with two hard drives.  Depending on which one I boot from, I get one of two different operating systems: both are windows 7, but they have different static IPs, serve totally different purposes, etc.    The workstation has a modern asus motherboard, core i7 intel CPU, 32 GB ram.   Is there a way to remotely change which hard drive is booted from on a reboot?  Both systems have remote users, and i'd like a solution that is more elegant than being physically present and selecting boot order manually. My HP Pavilion laptop has Ubuntu 18 version and I once tried to disable InBuilt webcam just for security reasons. But now it wasn't working and I reinstalled newer version of Ubuntu 20. But even after installing newer version the camera is still not working. When I open cheese it shows this screen. I assumed I might have deleted the drivers as I was doing package installation and removel that time. Please suggest me what should I do to get this working. Why does println(array) have strange output? ("[Ljava.lang.String;@3e25a5") How to wait in a bash script for several subprocesses spawned from that script to finish and return exit code !=0 when any of the subprocesses ends with code !=0 ?  Simple script:  #!/bin/bash for i in `seq 0 9`; do   doCalculations $i &amp; done wait   The above script will wait for all 10 spawned subprocesses, but it will always give exit status 0 (see help wait). How can I modify this script so it will discover exit statuses of spawned subprocesses and return exit code 1 when any of subprocesses ends with code !=0?  Is there any better solution for that than collecting PIDs of the subprocesses, wait for them in order and sum exit statuses? Accidentally moved directories under root Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n&gt;0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$  This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$  I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$  It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$ or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$ So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here). Show that $\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}$ I'm self studying from Category Theory in Context by Emily Riehl and have encountered the following question regarding slice categories:     Let $\text{C}$ be a category and $c$ be an object in $\text{C}$. Show that $\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}$    The exercise appears in an chapter that introduces the notion of opposite categories and duality proofs.     My intuition of why the statement is true comes from the following observation:  Let $x,y$ be objects in $\mathbf{C}$ (or equivalently $\mathbf{C}^{\text{op}}$). Let $g^{\text{op}} : c \to y$ and $f^{\text{op}} : c \to x$ be objects in $c/(\mathbf{C}^{\text{op}})$. By the definition of slice category, there exists a morphism $h^{\text{op}} : y \to x$ in $\mathbf{C}^{\text{op}}$ such that the diagram below commutes:  $\hskip2in$   Now the 'corresponding diagram' in the opposite category $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$ would be:  $\hskip2in$    which, as I understand it, would also be commutative. Now, to me, this indicates:   that the objects in $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$ are the same as the objects in $\mathbf{C}/c$, namely, a morphism in $\mathbf{C}$ from an object $z \in \mathbf{C}$ to $c$. the commutativity of the seconds diagram describes morphisms $h$ in both $\mathbf{C}/c$ and $(c/(\mathbf{C}^{\text{op}}))^{\text{op}}$ if you reverse the arrows again you return to the same diagram. This makes me think you could construct a map between morphisms by first making a map from $ \mathbf{C}/c \to (\mathbf{C}/c)^{\text{op}} ( \cong c/(\mathbf{C}^{\text{op}}) )$ by $h \mapsto h^{op}$     Now, what I am really looking for is a way to structure a proof. I feel like I have some intuition but am very unsure of how to write it down in a concise way. I am new to category theory so I'm still very much learning how to construct these types of proofs using duality; I feel like what I have written is maybe overkill but I'm trying to be as explicit as possible to make sure I understand everything.  In particular, I don't know how I would get a concrete correspondence between morphisms in the two categories that respects associativity of composition etc.  Any help or pointers would be appreciated! Skype window problem in Ubuntu 14.04 On my work computer I have two USB hard drives that I use rarely. They have a power save mode that sends them into sleep after a couple of minutes of them being idle.   Whenever I open a context menu on a file, the drives are woken up (most likely caused by the "send to" handler).  So I eject the drive, but I can't find a way to get it back, other than unplugging and replugging it in.  Is there a way to unmount the drives, and then remount them only when I actually need them? (On Windows 7 Ultimate.) I want to burn a Windows ISO to a USB device in Ubuntu. How do I do this?   I know how to burn a Ubuntu ISO into a USB device, but with a Windows ISO it's not the same. Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.   $"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$  then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$   stuck at this spot,  $(\sum_{n=0}^{\infty} a_nx^n)(\sum_{n=0}^{\infty} b_nx^n) = \sum_{n=0}^{\infty}(\sum_{k=0}^{n}(a_kb_{n-k}))x^n$  A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n&gt;0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.  EDIT  Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$ Get number of unread sms Suppose we want to define a first-order language to do set theory (so we can formalize mathematics). One such construction can be found . What makes me uneasy about this definition is that words such as "set", "countable", "function", and "number" are used in somewhat non-trivial manners. For instance, behind the word "countable" rests an immense amount of mathematical knowledge: one needs the notion of a bijection, which requires functions and sets. One also needs the set of natural numbers (or something with equal cardinality), in order to say that countable sets have a bijection with the set of natural numbers.  Also, in set theory one uses the relation of belonging "$\in$". But relation seems to require the notion an ordered pair, which requires sets, whose properties are described using belonging...  I found the following in Kevin Klement's,  (pages 2-3).  "You have to use logic to study logic. There’s no getting away from it. However, I’m not going to bother stating all the logical rules that are valid in the metalanguage, since I’d need to do that in the metametalanguage, and that would just get me started on an infinite regress. The rule of thumb is: if it’s OK in the object language, it’s OK in the metalanguage too."  So it seems that, if one proves a fact about the object language, then one can also use it in the metalanguage. In the case of set theory, one may not start out knowing what sets really are, but after one proves some fact about them (e.g., that there are uncountable sets) then one implicitly "adds" this fact also to the metalanguage.  This seems like cheating: one is using the object language to conduct proofs regarding the metalanguage, when it should strictly be the other way round.  To give an example of avoiding circularity, consider the definition of the integers. We can define a binary relation $R\subseteq(\mathbf{N}\times\mathbf{N})\times(\mathbf{N}\times\mathbf{N})$, where for any $a,b,c,d\in\mathbf{N}$, $((a,b),(c,d))\in R$ iff $a+d=b+c$, and then defining $\mathbf{Z}:= \{[(a,b)]:a,b\in\mathbf{N}\}$, where $[a,b]=\{x\in \mathbf{N}\times\mathbf{N}: xR(a,b)\}$, as in  question or  on Wikipedia. In this definition if set theory and natural numbers are assumed, then there is no circularity because one did not depend on the notion of "subtraction" in defining the integers.  So my question is:     Question Is the definition of first-order logic circular?   If not, please explain why.   If the definitions are circular, is there an alternative definition which avoids the circularity?   Some thoughts:   Perhaps there is the distinction between what sets are (anything that obeys the axioms) and how sets are expressed (using a formal language). In other words, the notion of a set may not be circular, but to talk of sets using a formal language requires the notion of a set in a metalanguage. In foundational mathematics there also seems to be the idea of first defining something, and then coming back with better machinery to analyse that thing. For instance, one can define the natural numbers using the Peano axioms, then later come back to say that all structures satisfying the axioms are isomorphic. (I don't know any algebra, but that seems right.) Maybe sets, functions, etc., are too basic? Is it possible to avoid these terms when defining a formal language? I can't type a “b” when I'm in the command line Conditional distributions of the multivariate normal Every element of a group has order $2$. Why, intuitively, is it abelian?