Algorithm to decide if a CTL formula can also be expressed in LTL and translate it?
Syntactic characterisation of the intersection of CTL and LTL
While reviewing First Posts on Super User, I noticed that I couldn't edit this question (see below). Clicking the &quot;edit&quot; link simply does nothing. Editing is not possible for me right now because I have already five suggested edits pending. It used to be the case that the link would've been grayed out (and the reason shown as a popover). , and while I'm not sure it's an improvement, the current implementation needs some fixing at least.  This bug appears regardless of the actual reason that the edit isn't possible; for instance,  is another user experiencing the same bug who couldn't suggest an edit because of an already pending one. Apparently, I'm far from the only one who experiences this issue. As of last Sunday morning,  have been posted networkwide (and are still alive) and  is #16. Transforming Differential Equation to a Kummer's Equation I'm trying to transform an equation of the form $$ yw^{\prime\prime}(y) - [b - ay] w^\prime(y) - [d + ey]w(y) = 0 $$ into the form of a Kummer's or confluent hypergeometric differential equation: $$ y w^{\prime\prime}(y) + [f - y] w^\prime(y) + g w(y)  = 0 $$ I know it may have something to do with merging two of the singularities of the original equation, and maybe doing something with $y$, making it $\frac{y}{b}$ and taking b to infinity, but I don't know and can't find the details for this process, and for my equation specifically. There's not too much difference between the two but just enough so that I can't get it. Thanks in advance for any help! Cancelling special & general relativistic effects If $Ax = Bx$ for all $x \in C^{n}$, then $A = B$. Let $A$ and $B$ are $n\times n$ matrices and $x \in C^{n}$.  If $Ax = Bx$ for all $x$ then $A = B$. To prove this I have selected  $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}. Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$. Hence $A = B.$    Is this proof complete? How do I use a different drive as an installation location for Steam? I recently got an invitation for Steam. I want to download games but my SSD drive which I have Ubuntu will not be enough. How can I install Steam on a different hard drive?  I went into the "Steam Library Folders" but I can't delete the original folder so steam can install on that drive. I have also tried to change the place but it is again on the down side: Proving that the set of $\nu$ measurable sets $M\subseteq P(X)$ is an algebra Why does prime factorization for LCM work? A method to finding the LCM of two numbers is to use prime factoring. I know how to do this, but logically why does this work in finding the lcm What does the function f: x ↦ y mean? I am doing , and am having a few problems with function notation. I understand the form f(x).  What does the form f: x ↦ y mean? Could you also give one or two examples?  And, if possible, state your source. Thank you. Backup with duplicity on Google Cloud Storage I have already set a backup routine that backups all my files to Amazon S3 using duplicity. I would like to create a similar backup routine with Google Cloud Storage. Is it possible to get duplicity to work with GCS? Are smaller p-values more convincing? Can I bring my laptop, camera on flight to USA? I am going to the US in the coming week. However, I heard that laptop and camera are not permitted to bring on board as carry-on. What are the latest rules about these larger electronics? Let $X$ be a topological space. The diagonal of $X \times X$ is the subset $$D = \{(x,x)\in X\times X\mid x \in X\}.$$   Show that $X$ is Hausdorff if and only if $D$ is closed in $X \times X$.   First, I tried to show that $X \times X \setminus D$ is open using the fact that $X \times X$ is Hausdorff (because $X$ is Hausdorff), but I couldn't find an open set that contains a point outside $D$ and is disjoint to it... Creating intuition about Laplace & Fourier transforms Center Gravity in Unity a 2d planet in the center, where the players will be able to walk on it and also will be affected by center gravity. planet will be static and there won't be any other planets. so;  Is there a way that I can change the gravity position? yes you can adjust gravity via project settings but I want it to be in the middle x,y,z(0,0,0)  I have tried creating my own gravity but it wasn't very useful because objects wouldnt fall, stand on foot and also in contact with another object they would vibrate instead of standing still. (if they stand on the planet they would vibrate)          private void FixedUpdate() {     foreach(GameObject item in grav)     {         Attact(item);     } }  private void Attact(GameObject gmf) {     Rigidbody2D rgg = gmf.GetComponent&lt;Rigidbody2D&gt;();      Vector3 direction = rg2d.position - rgg.position;     float distance = direction.sqrMagnitude;      float forceMag = (10 * rgg.mass) / distance;     Vector3 force = direction.normalized * forceMag * 10;      rgg.AddForce(force); }   I can create a script that will change the gravity by the opposite of the object's position. this does work but I cannot see any way to do it for more than one object. Limits, find a limit that exist in absolute value but not outside the absolute value. How can I use an array modifier to create individually manipulatable objects? I want to know the correct form:     He and I will go there   or   He and me will go there.   Which is correct? Accepted answer appearing below top voted On , sorted by votes, the accepted answer is #3 in order of votes.  It appears as number 3 in the list as well. Is this new behavior intended? How does Newtonian mechanics explain why orbiting objects do not fall to the object they are orbiting? I understand there is a method for solving simultaneous modular equations. For example; $$x = 2 \mod{3}$$ $$x = 3 \mod{5}$$ $$x = 2 \mod{7}$$ We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by: $$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$  But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x?