I need to prove that the ∃x(R(x)→∀yR(y)) is logically valid. I'm trying to understand why this statement is true but I can't figure it out.
Why is this true?  $(\exists x)(P(x) \Rightarrow (\forall y) P(y))$
This doesn't work:  tar xf /tmp/foo.tar.gz foo/bar tar: foo/bar: Not found in archive   It's not obvious to me what would do this beyond extracting it in place and moving the files over. Vertical spaces between lines in cases environment \documentclass{article} \usepackage{amsmath} \begin{document} \begin{equation*}   X_i =    \begin{cases}     \left[\begin{smallmatrix}       0 \\ 1 \\ 2     \end{smallmatrix}\right]     &amp;     \text{if $i$ is even;}     \\     \left[\begin{smallmatrix}       2 \\ 1 \\ 0     \end{smallmatrix}\right]     &amp;     \text{otherwise.}   \end{cases} \end{equation*} \end{document}   This is the LaTeX code that I wrote based on a cases environment. There's no space between the matrices on the right. What's the best way to add some vertical spaces to separate them?    p.s. This is a simplified form of my original writing, so I cannot just accept suggestions like 'use a matrix transpose to transform the column vector into a row vector.' Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2 What is the command to remove all files but not directories? Let's say I have a directory tree like this:  FOLDER:     file1     file2     file3     Subfolder1:         file1         file2     Subfolder2:         file1         file2   If I used rm -r FOLDER/*, everything in FOLDER would be deleted including sub-directories. How can I delete all files in FOLDER and in its sub-directories without deleting actual directories? How to express a polar coordinate for a macro that only accepts space-separated values? \psElectricfield only accepts a coordinate in space-separated values as follows. For example, a charge of -0.1 unit at (-1,0) is written as [-0.1 -1 0]. I have attempted to write in PS notation but it still does not work.  \documentclass{article} \pagestyle{empty}  \usepackage{pst-electricfield,multido} \def\M{2}  \begin{document} \multido{\i=0+30}{12}{% \begin{pspicture*}(-\M,-\M)(\M,\M)     \psElectricfield     [         Q={[-0.1 -1 0][-0.1 1 0][0.1 !\i sin 2 mul \i cos 2 mul]}     ] \end{pspicture*}\newpage} \end{document}   How to express  a polar coordinate for a macro that only accepts space-separated values? In the case of a non-inverting amplifier, if I DC couple the input signal with a capacitor only, \$C_1\$ (i.e. no resistor to ground), sources (as well as an experiment in the lab) show me that the op amp input will be saturated as the input bias current deposits charges on the terminal without having a return path to ground. This is solved by using connecting a resistor (\$R_1\$, as shown in the image below) to ground to create the return path. [The image is reproduced from an answer by "Neil_UK" on one of my previous questions, thanks Neil!]    My questions:   1) How is the input bias current able to continuously deposit charges and saturate the input if I don't use the resistor? The resistor should not allow the DC bias current through, so where is this current flowing from/into?  2) Why is this not a problem for the feedback path to the "-" terminal? There is no resistor to ground in that case, and capacitor \$C_2\$ seems to be causing no issues. The PDFs I've been reading suggest that the feedback behaves similar to when I have a resistor to ground by acting as the return path, how is this? Event binding on dynamically created elements? I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option. ggplot2: facet_wrap strip color based on variable in data set I want to play a game in an older update, but can't see a way to choose a specific version. Is this possible? I have to do some manual reordering or rows in Excel and cut/paste is too hard to use. I'm looking for something that would enable me to use drag'n'drop or to add some buttons to move-up/down move-top/bottom. $|f(x)-f(y)|\geq k|x-y|$.Then $f$ is bijective and its inverse is continuous. My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!! Get rid of the question ban on Meta; suspend actively harmful users manually Show that quotient rings are not isomorphic I've been given a homework problem that requires me to show that the rings $\mathbb{C}[x,y]/(y - x^2)$ and $\mathbb{C}[x,y]/(xy-1)$ are not isomorphic.    This is my attempt at a solution:  For $\mathbb{C}[x,y]/(y - x^2)$, we can parametrize in the following way: $x = t$ and $y = t^2$. Then this ring is isomorphic to $\mathbb{C}[t]$.  For $\mathbb{C}[x,y]/(xy-1)$, we can parametrize $x = t$ and $ y = 1/t$. Then this is isomorphic to $\mathbb{C}[t, 1/t]$.   But $\mathbb{C}[t, 1/t]$ is not isomorphic to $\mathbb{C}[t]$.  Am I on the right track? If not, any helpful hints? I have multiple cameras in a scene. What I want to do is to change which one is active. This way I can view the scene from different angles. How would I do this? How can orbitals be oriented if the nucleus is symmetrical? I know very little about quantum physics, but I just can't wrap my head around the fact that non-spherical orbitals appear to be "oriented", constituting a set of fixed "frame of reference axes".  Does it really work like that? Are there for a single atom given directions in which the orbitals "protrude"? And if so, what are they based on? If I understand it right, the nucleus is seemingly spherical - symmetrical with no unique axis of symmetry. What would then decide the orientation of the orbitals in space (with respect to the surroundings)?  Edit: I understand any orientation works to satisfy the wave equation. I was more curious why the atom would choose one direction, instead of any of the others. Is that fundamentally random? And is this setup retained further, or can it change at any time? How do I prove or disprove $X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)$  for all sets $X$, $Y$, and $Z$? I'm lost on the steps here. Meaning of syntax $N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ (multivariate normal distribution) MSO Review Queue Privileges are Misleading When I visit Stack Overflow, I see the review option in the bar at the top of the screen.    However when on MSO, I don't see it. I assumed this was because I didn't have enough reputation to access the queue. I was curious what the requirement was on meta, so I moseyed on over to the privileges page. To my surprise, I saw that the privilege is listed as only requiring 500 rep and informs me I already have it:    Confused, I manually typed in meta.stackoverflow.com/review to see what would happen. The page did in fact load, but as you can see meta has no "First Posts" or "Late Answers" queue.    Seeing on that page that a minimum of 2k reputation is required to access any queues on MSO, I report the following bug:  1) On the privileges list on MSO, the 500 rep privilege of "Access first posts and late answers review queues" should not appear.  I also have the following feature request:  2) On the privileges list on MSO, it would be useful if instead there was a 2k privilege that informed you that you have access to the "Suggested Edits" and "Low Quality Posts" queues. Why did Dumbledore make it less difficult to get to the Sorcerer's Stone? The Sorcerer's/Philosopher's Stone was removed from Gringotts to be placed under the most impenetrable of protections at Hogwarts under Dumbledore's custody. But it seems clear Dumbledore took protections which could have been impenetrable, and deliberately made them less so, such that at best they would only slightly slow down anybody trying to access the Stone, rather than just plain block them.  For example, after the Devil's Snare, you have a locked door enchanted such that it cannot be opened by any charm or other means except its own assigned key. That's a great protection, potentially an impassable barrier if the key is not available.  But the key is then left right there in front of the door in the same room.  At least the key is flying around the ceiling. But there are broomsticks right there in the same room, too.  The room with the fires front and back—that could also have been a solid defense—deadly fires that cannot be passed through unless you drink very specific potions tailored to each particular fire's properties.  Except that those potions providing safety are put right there in the same room.  At least, perhaps, they're surrounded by other bottles, some of which contain poison. With no way to know which bottle is safe, that could still almost be a solid protection.  Except that a piece of paper with a logic puzzle detailing exactly which bottles are which is put on the table with them.  Dumbledore always had reasons, and those reasons were always virtuous. What was he up to, deliberately weakening the protections in this way? If angular velocity & angular acceleration are vectors, why not angular displacement? Are angular quantities vector?   ... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction of the vector. In the world of pure rotation,a vector defines an axis of rotation, not a direction in which something moves.   This is what my book (authored by Halliday, Walker, Resnick) says. But they mentioned with a caution that      Angular displacements cannot be treated as vectors. ....to be represented as a vector, a quantity must also obey the rules of vector addition..... Angular displacements fail this test.   Why do they fail in this test? What is the reason?