Why is p ∧ q ⇒ r true when p is true and q is false?
In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?
Write to FIFO only if it exists I need to find the closure of these sets on the ordered square: $$A = \{(1/n)\times 0 | n\in \mathbb{Z}_+\}$$ $$B = \{(1-1/n)\times \frac{1}{2}| n\in \mathbb{Z}_+\}$$ $$C = \{x\times 0 | 0&lt;x&lt;1\}$$ $$D = \{x\times \frac{1}{2}| 0&lt;x&lt;1\}$$ $$E = \{\frac{1}{2}\times y | 0&lt;y&lt;1\}$$ I know that the closure is just the set itself united with its limit points. For $A$, the set of limit points, I imagine, are $\{&lt;1,0&gt;\}$, that is, the $x$ coordinate is $1$ and the $y$ is $0$. But  is says the reverse. What am I getting wrong? Following the same reasoning, $B' = \{&lt;1,\frac{1}{2}&gt;\}$ $C' = \{&lt;0,0&gt;,&lt;0,1&gt;\}$ What am I doing wrong? Prove that 1/2 + 1/4 + 1/8 ....... = 1 I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.  It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations. How to print original variable's name in Python after it was returned from a function? I have enum and use the variables like myEnum.SomeNameA, myEnum.SomeNameB, etc.  When I return one of these variables from a function, can I print their names (such as myEnum.SomeNameA) instead of the value they returned? What can $E=mc^2$ do? In the famous equation , the variables stand for:     $E$ is energy, $m$ is mass, and $c$ is the speed of light (in vacuum).   And I understand the equation fairly but limited in knowing in its effects/outputs in reality, honestly.   Some example? Could anybody please give me some sort of examples in real world which are done by this equation? How do I determine if an array is initialized in VB6? Passing an undimensioned array to the VB6's Ubound function will cause an error, so I want to check if it has been dimensioned yet before attempting to check its upper bound.  How do I do this? Deploy a project using Git push Is it possible to deploy a website using git push? I have a hunch it has something to do with using  to perform a git reset --hard on the server side, but how would I go about accomplishing this? Strange error defining a macro with arguments for math mode How can I add a bookmark in my nautilus folder explorer Find all functions $ f : \Bbb R \to \Bbb R$ which satisfy $f(x)f(y) = f(x + y) + xy$ for all $x, y ∈ \Bbb R$. Find all functions $f : \Bbb R \to \Bbb R$ which satisfy $f(x)f(y) = f(x + y) + xy$ for all  $x, y \in \Bbb R$. I was loading an image resource with the following code and it works fine from my dev environment (netbeans) but threw an exception as soon as it ran from the desktop.  new ImageIcon(Images.class.getResource("images/highlighter24.gif"));   I traced it down to my file having different case than in the code.  Fine, my bad, but why the inconsistent behavior?  Gremlins?  Note: I'm running on windows before you ask. Stern-Brocot tree where each node has two fathers How to add an image to a JPanel? I have a  to which I'd like to add JPEG and PNG images that I generate on the fly.  All the examples I've seen so far in the , specially in the  use ImageIcons.  I'm generating these images as byte arrays, and they are usually larger than the common icon they use in the examples, at 640x480.   Is there any (performance or other) problem in using the ImageIcon class to display an image that size in a JPanel? What's the usual way of doing it? How to add an image to a JPanel without using the ImageIcon class?   Edit: A more careful examination of the tutorials and the API shows that you cannot add an ImageIcon directly to a JPanel. Instead, they achieve the same effect by setting the image as an icon of a JLabel. This just doesn't feel right... Best explanation of Ruby blocks? Recovering from a move of /home when it's encrypted (ecryptfs) I had a 20GB / partition at work that dual-booted with Windows and it quickly ran out of space. I usually do 20GB / and 160GB (or whatever the rest of the hard drive is) as /home, so I intended on recovering the space Windows was taking up and turning that into /home.  That went fine and I was able to mount the partition and rsync all my files over.  However, I forgot that the partition was encrypted, so now all my files are sitting in /home/.ecrypt and /home/$USER is empty.  How can I recover my files from /home/.ecrypt and move them to /home/$USER?  I followed this tutorial: Get the image URL I want to render an image as a background image via an inline style in twig. I created a field called bg_image and attached it to the standard plain page.  After fiddling around for hours I was able to get the Image URL in node.html.twig  {{ file_url(node.field_bg_image.0.entity.uri.value) }}   but I could not get it to work inside field--field-bg_image.html.twig  Can I get the node from there anyhow, so I can then get the image?  How can I get the image URL to use as inline style? I thought maybe I can pass a variable from field--field-bg_image.html.twig to image.html.twig and then just render  {{ uri }}   instead of  &lt;img{{ attributes.addClass(classes) }} /&gt;   but I could not get it to pass the variable there unless I use includes  {% include 'image.html.twig' with {'image': image, 'isFromField': isFromField} %}   (isFromField is true, when it comes from the field--field-bg_image.html.twig) But that didn't work either. The Image was never rendered that way.  Would be very glad if you can help - my php knowledge is very basic. Thanks Ugly "Next Privilege" text on user activity page How can I get the amount of available memory portably across distributions? The standard files/tools that report memory seem to have different formats on different Linux distributions. For example, on Arch and Ubuntu.   Arch  $ free               total        used        free      shared  buff/cache   available Mem:        8169312     3870392     2648348       97884     1650572     4110336 Swap:      16777212      389588    16387624   $ head /proc/meminfo  MemTotal:        8169312 kB MemFree:         2625668 kB MemAvailable:    4088520 kB Buffers:          239688 kB Cached:          1224520 kB SwapCached:        17452 kB Active:          4074548 kB Inactive:        1035716 kB Active(anon):    3247948 kB Inactive(anon):   497684 kB  Ubuntu  $ free              total       used       free     shared    buffers     cached Mem:      80642828   69076080   11566748    3063796     150688   58358264 -/+ buffers/cache:   10567128   70075700 Swap:     20971516    5828472   15143044   $ head /proc/meminfo  MemTotal:       80642828 kB MemFree:        11565936 kB Buffers:          150688 kB Cached:         58358264 kB SwapCached:      2173912 kB Active:         27305364 kB Inactive:       40004480 kB Active(anon):    7584320 kB Inactive(anon):  4280400 kB Active(file):   19721044 kB    So, how can I portably (across Linux distros only) and reliably get the amount of memory—excluding swap—that is available for my software to use at a particular time? Presumably that's what's shown as "available" and "MemAvailable" in the output of free and cat /proc/meminfo in Arch but how do I get the same in Ubuntu or another distribution? I am confused by the orientation of a topological manifold. My understanding is: An orientation of a topological manifold is a choice of generator of the $H^n(M,\mathbb Z)$. So given a manifold, we could have 2 orientation defined on the manifold. For $\mathbb{CP}^2$ on orientation is determined by the complex structure, the other orientation is denoted by $\overline{\mathbb{CP}}^2$. And it is well known that there is no orientation reversing map from $\mathbb{CP}^2$ to itself.  My question is: are they homeomorphic? I guess they are not homeomorphic. But I am confuesed, doesn't the orientations defined on the same manifold, how come after reversing the orientation they become not homeomorphic? Genetic linkage greater than 50 centimorgans Classically, the linkage between two loci can be measured in centimorgans (cM), which represents the percent chance that these two loci will recombine an odd number of times (generating a recombinant genotype).  Due to independent assortment, markers on different chromosomes are expected to recombine 50% of the time. As I understand, if one conducts a classic experiment and counts various progeny, then deduces the apparent linkage, values of 50 cM or more are interpreted as "impossible to determine whether on separate chromosomes or just recombines very often". See (), with grammar corrected:     The final point that we need to make regards the maximum distance that we can measure. Because of the way in which the calculations are performed, we can never have more (than) 50% recombinant gametes. Therefore the (maximum) distance that two genes can be apart and still measure that distance is just less (than) 50 cM. If two genes are greater than 50 cM apart, then we can not determine if they reside on the same chromosome or are on different chromosomes.   Wikipedia gives an  and also remarks that (d is physical distance):     The probability of recombination is approximately d/100 for small values of d and approaches 50% as d goes to infinity.   However, what happens if the linkage is greater than 50 cM? It seems like classical experiments aside, such a situation is possible in reality.  For instance, , and . Let's take two genes on Chr.IV:    is at about +631 kb.  is at about +1370 kb.   The physical distance between these is about 734 kb. According to the cM/kbp value, the centimorgan distance is $734\text{ kb} \cdot 0.31\text{ cM}/\text{kb} = 228 \text{ cM}$ . How to interpret this 228?  If I mate a haploid yeast which is DNF2 TOM1 with another which is dnf2 tom1 (lowercase indicating minor allele rather than deletion), then sporulate them, what is the chance of getting spores with DNF2 tom1 or dnf2 TOM1 genotypes?  I realize that the cM/kbp value is only a simplification and that in practice linkage is a more complex phenomenon. Nevertheless, it seems plausible that several crossover events can happen on Chr.IV since it is so big. This includes the possibility of 1, 3, 5 and more crossovers which would create a hybrid (assuming they all happen between these two loci), as well as the possibility of 2, 4, etc crossovers which would create a non-hybrid spore (at least as far as our chosen markers are concerned).