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Can I successfully bake previously-frozen bread dough? If $\Gamma$ is satisfiable then $\Gamma$ is consistent We say that $\Gamma$ is consistent if for no $\phi$, then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$. We say that $\Gamma$ is satisfiable if there is a structure $M$ s.t. $\models_M\gamma[s]$ for alle $\gamma\in\Gamma$.  I want to show that if $\Gamma$ is satisfiable, then $\Gamma$ is consistent.  I'm thinking we assume that $\Gamma$ is not consistent, and then show that $\Gamma$ is not satisfiable. So if $\Gamma$ is not consistent then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$ for some $\phi$. But by soundness theorem then $\Gamma\models\neg\phi$ and $\Gamma\models\phi$, meaning for structure $M$, we have $\models_M\neg\phi[s]$ (whereas $\not\models_M\phi[s]$) and $\models_M\phi[s]$. So would this mean that $\Gamma$ is not satisfiable? I'm a little confused, because wouldn't this instead be a contradiction? I know this may be a tough or confusing question, but I have numerous existing materials I created and reuse for a project, which were all prepared for Blender Internal.  Now, I want to start the transition to Cycles, but I've hit a roadblock. How can I transition my materials without completely starting over? Or should I just forget it and start with fresh materials? The locus of the intersection point of two perpendicular tangents to a given ellipse For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.   I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof. Will my Chinese visa still be valid if I renew my passport? Will my Chinese visa still be valid if I renew my passport? If so how does that work?  I have heard that you can travel with your new passport and just show the old passport with the valid visa when entering China. Has anyone done that? Recently?   I have also been told that you have to get a new visa, which seems somewhat logical but frustrating, especially when my Chinese visa is only a few months old.  My Chinese visa was issued in late Nov 2014 and I heard the rules changed between the US and China. My visa says it's good for 10 years and my US passport expires in 5. variable-sized "such that" pipe Can this statistic be shown not to be sufficient for $\theta$? This problem comes from , who do not rigorously demonstrate (in their solution key) that the statistic is not sufficient.  Let $X_1,\dots,X_n$ be a random sample from a population with PDF $f(x|\theta)=\theta x^{\theta-1} \cdot 1_{\{x\in(0,1)\}}$ for $\theta&gt;0$. Show that $\sum_i X_i$ is not sufficient for $\theta$.  If you write out the PDF $p(\vec{x}|\theta)$ of the random sample, it is clear by the factorization theorem that $\prod_i X_i$ or $\sum_i \log(X_i)$ are sufficient for $\theta$; the PDF $p$ also suggests $\sum_i X_i$ is not sufficient for $\theta$. However, to rigorously demonstrate this, we need to analyze $p(\vec{x},\theta)/q(T(\vec{x},\theta)$, where $p$ is the distribution of sample $\vec{x}$, $q$ is the distribution of the statistic $T(\vec{X})=\sum_i X_i$. But finding the distribution of $T(X)=\sum_i X_i$ seems impractical. I have observed that $f$ is the PDF of a Beta($\theta$,1) distribution, but checking online, the distribution of the sum of Beta random variables doesn't appear to have a closed form. Are there any alternative routes (e.g., showing there is no factorization involving $T(\vec{X})$)? Did C&amp;B leave out a full explanation that $\sum_i X_i$ is not sufficient because none actually exist? Why are RSA key sizes almost always a power of two? I know that other bit sizes are possible, e.g. this HTTPS server seems to have a 9000 bit key , but it's very rare that one sees a key not of size 1024, 2048, 4096, etc. bits in common usage - what is the reason for this? Is it just customary, or is there a cryptographic advantage? What is the view on accounts shared by multiple users on Stack Overflow? It has been my understanding that accounts are suppose to be for a specific user. This user, , is a company: . No doubt, this account has provided some quality answers. However, if a specific user cannot be identified, I don't think that the rep gain, rewards, or badges should be given. This should somehow be tagged as a 'Corporate User'. The rep has no real value when you can't be somewhat sure that the same person is giving the answers all the time.  This is where I found an answer by Devart: , peddling a product.    Edit 2 Good points have been brought up and most I can agree with. But we are setting a precedent. Meta Fudgey brought up a good point about the account's ability to use mod tools responsibly and consistently. Maybe this feature is not available to an account that appears to be an umbrella account unless a single user is identified. I think it is important to consider this aspect.    Let me re-iterate: the user(s) has provided excellent content and I would want that to continue. However, there is more at stake here than meaningful content and helpful resources.      Edit The view on 'shared' accounts wasn't so bright a year ago. For most of us, we have information in our profiles that identifies us as individuals. No, I don't have absolute knowledge of the activities on this account and how many people might use it. Meh, for all we know it could be a knowledgeable programmer with some good PR sense. On the other hand, it could be a PR person with no programming knowledge getting the answers from an employee. Who knows... What happens to the second Marty at the end of Back to the Future? In  (the original movie), Marty comes back from 1955 10 mins early to prevent Doc from being shot (which he fails at). He then rolls up to the Lone Pine Mall and sees himself hit 88 mph and shoot back to 1955 (again). Marty then carries on with this life as usual. What happens to the Marty that goes back to 1955 at the end of the film?  This obviously is a recursive problem where it would create another timeline and Marty would keep coming back 10 mins early to view another Marty zip back to 1955? Can anyone suggest a detailed installation process for Ubuntu on Windows 8.2? Can QGIS preserve layers when exporting to PDF? How can I find $\lim_{n\to \infty} a_n$ I am trying to solve the following:  Let $W_t$ be a brownian motion and $a,b &gt;0.$  Define $\tau$ to be the exit time of $W_t$ from $[-a,b]$, that is $$\tau = \inf\{t \ge 0\ :\ W_{t}(\omega) \notin [-a,b]\}.$$  Show that $\tau$ is integrable and compute $\mathbb{E}(\tau)$.  I have seen this stated without proof multiple places that $\mathbb{E}(\tau)=ab.$  What is the logical argument that backs up this statement?  How are we guaranteed $\tau$ is integrable? Relation between power and energy If I have an expression for the 'rated generator power' of a wind turbine, call it $P(t)$ how is the energy, that the wind turbine generator generate, related to the power $P(t)$?  So: $$E(t)=\dots$$ Find all Quotient (or Factor) Groups of $D_4$ (Dihedral Group of order $8$) I need to be able to find all of the quotient groups for dihedral group 4 with $D_4 = \{ e,R,R^2,R^3,V,H,D,D'\}$. I know I have to start by finding the normal subgroups, which are $\{e,R^2\}$ $\{e,R,R^2,R^3\}$ $\{e,R^2,V,H\}$ $\{e,R^2,D,D'\}$. Then I need to find the sets defined by $G/H=\{ aH : a \in G\}$ with operation $aH  bH = abH$. I am stuck here; could someone please show me how to find all of the factor groups for D4? The Maximum possible order for an element $S_n$ Multiply cited reference in footnotes Is it possible to run some code when an assembly is loaded, without doing anything specific in the loading code? What I am looking for is like a static constructor on a type.  Eg:  Assembly A does not know about Assembly B, but B does know about A. Assembly A needs to know certain things about B if B is loaded. When Assembly B is loaded by the runtime (referenced, or explicit), I want a piece of code (static method or attribute) to be executed that calls a method in Assembly A.  The root cause of this problem is unknown types being encountered when serializing a type in A that contains types from B not known at compile time as interfaces are used. I have a couple of eggs that I've picked up from some Pokestops (I'm lucky enough to live above one).   I walk a lot. I get to work through a 2 mile walk / train journey every day, so I should be picking up loads of steps for the eggs I have. However, because I'm normally always running late... for everything, I have to rush everywhere, so I walk at an above average pace.   Today I've walked around 5 miles, but the game has registered 100 metres. Is this because I walk too fast, or because of something else?  It's worth pointing out that I have the app open when walking... mostly because I have to catch them all. (This made me miss a train this morning... luckily I was early :D)