Trigger outbound message/API Endpoint on record delete
Why Process Builder doesn't work on Delete Scenario?
When & why I need to use web.AllowUnsafeUpdates = true Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$ Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.  I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question. Let $\{E_n\}_{n\in\mathbb{N}}$ be a sequence of countable sets and let $S=E_1\times\cdots\times E_n\times\cdots $. Show $S$ is uncountable. Prove that the same statement holds if each $E_n=\{0,1\}$.  By the definition of Cartesian product of sets,   $$\displaystyle S=\Pi_{n\in\mathbb{N}} \{f\colon\mathbb{N}\rightarrow\cup_{n\in\mathbb{N}}E_n\mid\forall n, f(n)\in E_n\}$$  If $E_n=\{ 0,1\}$, then  $$\displaystyle S_{01}=\Pi_{n\in\mathbb{N}}\{0,1\}=E^{\mathbb{N}}$$, where $E=\{0,1\}$.  By a theorem, $\cup_{n\in\mathbb{N}} E_n$ is countable since the sequence is countable.  I'm not sure how to go on from here to show $S$ is uncountable. Can we say anything about the function $f$ that maps a countable $\mathbb{N}$ to another countable union of sequence of countable sets? How can I approach implementing the following in Drupal 7?  What I need to do is create a module which defines a new fieldable entity, called 'Company'. I have a list of, let's say, 20 fields that need to be filled out by each Company instance. These questions are predefined and some may contain custom validation.  At the moment, I'm at the point when I can add new fields to the Company entity. This works fine at the moment. My issue is that I need all these fields to be there as soon the module is installed, so adding them via the interface isn't an option.  I was wondering how can I approach this? I'd assume it comes down being able to do what can be done using the 'Manage Fields' UI programmatically. It often occurs to me to add an edit summary to a question-edit immediately after submitting the edit, especially if the changes aren't obvious (formatting etc.). So I:   Click "edit" again. Fill in the summary Click on "Save Edits".   The 'edit' (no actual change has been made to the title/body/tags)  is accepted, but the summary doesn't show up, i.e. it is silently rejected.   I don't think it should work this way because the summary has been added by the original editor within the original edit's edit-window. In fact if I remember correctly, it to used to work as expected earlier; so perhaps this is a recent change. Badge icons missing in Achievements of mobile view The badge icon followed by the site icon in the Achievements dialog is missing in the mobile view. But the same is available in the full site view.  Screenshots for reference:  Achievements - Full site view:        Achievements - Mobile view: How to delete empty folder with thumbs.db without disabling thumbnails I often have problem to delete a folder that contains a thumbs.db file. Also tools like  fail. It seems to be loccked by os for some unknown reason. The only solution i found in the net was to disabling thumbnails in windows but this is a bad workaround isn't it?  Is there a "normal"(or official, supported by MS) way to delete an empty folder that only contains a thumbs.db? Where to ask Question that can be flagged too broad or opinion based How to change Facebook username? Do I have to re-register a new Facebook account to change my facebook.com/USERNAME?  Or are there any ways to change it? I have an object already created in blender, a tree, or a human, or a pipe.. and I need this to be drilled with multiple holes.   The holes are painted in an image, that is, it is an image that has a repeating an images pattern (they can be circles, squares, or any form).  Here are some examples for drilling-pattern: More informative flagging history for comments Need a way to do a "loop cut" on a triangle fan Need a way to do a loopcut- ish thingy on the blue line Commutator Subgroup and the Largest Abelian Quotient Good day. I'm not too sure how I should go about asking my question, so if it is in anyway confusing, feel free to edit it.   In my Group Theory course we defined and showed a few properties of the commutator subgroup $[G,G]$ of a group $G$. And then we stated and proved a lemma in lecture that I'm having some difficulty wrapping my head around. I'll state it here for clarity.  Lemma a) The commutator subgroup $[G,G]$ of $G$ is a normal subgroup of $G$. b) The quotient group $G / [G,G]$ is the largest abelian quotient of G in the sense that every homomorphism $\varphi : G \to A$ where $A$ is an an abelian group, induces a unique homomorphism $\bar \varphi : G/[G,G] \to A$ so that $\varphi =\bar \varphi \circ \pi$ where $\pi : G \to G/[G,G]$ is the quotient homomorphism.  Now, it's not the proof of this lemma that I'm having trouble with, it's the italicized text that is messing with me. I completely understand what it is that the italicized text is saying, I'm just not too sure why it is that the property gives us the "largeness" that is claimed. I can see that the above forces $G$ to "pass through" $G/[G,G]$ before it "reaches" $A$ (I apologize for the casual terminology), so I can sort of see why $G/[G,G]$ should be the largest quotient, in a non-rigorous sort of way.  Overall I'm very confused. Perhaps I'm missing something very simple. Any help or nudges in the right direction would be much appreciated.   As a side note I did try a search on Google and here on Stack Exchange and couldn't find anything that answered my question. So if I missed something obvious, then my apologies. Thank you in advance. What were Luke's lessons? Luke tells Rey that      there are three lessons he will teach her as he trains her as a Jedi   What are the lessons? (Disclaimer: I've read the  and I do understand that this is bordering on off-topic since it's about consumer devices. However I hope that I've managed to make enough on-topic by asking about why it is the way it is, and what is the explanation from an electromagnetic point of view - in other words, what laws of electricity and what practical considerations have gone into these decisions).  I recently bought a power tool and in its manual I found a few peculiar safety rules. I cannot see why they would be there, although reading up more it seems that they are commonplace. My understanding of electricity is limited (just about enough to be dangerous), so I'm hoping to expand it. :)  The rules are:   Don't use an extension cord longer than 25m. Considering this is an outdoors tool, this limitation is severely restrictive. The best explanation I've seen is that longer cables would simply have too high of a resistance and the tool wouldn't be able to operate at full power. But this argument doesn't convince me. The tool is rated at 2.4kW, and it's meant to be used with 220V AC power source, which means it uses about 11 amps of current. All the extension cords I've ever come across have at least 16A limitation, so there is quite a reserve still. In addition, the power has to travel much, much further to get to my house, and even after the meter it still takes probably close to those same 25m to get to the power outlet. Why only 25m after the outlet? The resistance of the cables isn't that big, is it? Don't daisy-chain extension cords. I suspect this is a variation of the above - don't exceed 25m; every connection adds additional contact resistance; water in the outdoors connections can be a problem - in general, it just adds risk. However, apart from the water in the contacts, the rest seems to me to be practically negligible. Is there really a significant contact resistance? Assuming that the extension cords are in a good shape. Make sure that the power cord is entirely unrolled. This is another odd one. My suspicion is that it has to do with inductive resistance. But is it really that strong? Wherever I've seen people actually wanting to create an electromagnet, it's pretty much hard work. You need a lot of neat, tightly-wound loops of wire. An extension cord has a lot of insulator around it, spacing the wires pretty far apart, and the loops are very chaotic. In addition, if that was a problem, why do  exist? They're used especially when you need a very long extension cord, and you don't need to unroll them completely every time you want to use them. Solved: OK, I was off the mark. The correct answer (also a ) has to do with the wire heating up from being used close to its limits. When coiled together, the temperature can go pretty high and melt the cables. Not fun. Let $f:X \to Y$ be a homeomorphism between topological spaces $X$ and $Y$. Let $A \subseteq X$ and $B = f(A) \subseteq Y$ (given the subspace topology) and let $g: A \to B$ be the restriction of $f$ given by $g(a) = f(a)$ for $a \in A$.  Showing that $g$ is a homeomorphism.   $g$ injective   Let $g(a_1) = g(a_2)$, $a_1, a_2 \in A$.  $\implies f(a_1) = f(a_2)$  $\implies a_1 = a_2$ as $f$ homeomorphic.   g surjective   Let $y \in B$  $\implies y = f(a)$ for some $a \in A$.  $\implies y = g(a)$ for some $a \in A$.   $g$ continuous   Let $U \subseteq B$ be an open set in $B$.  $f(a) = g(a)$ for all $a \in A =&gt; f^{-1}(b) = g^{-1}(b)$ for all $b \in B$.  So $g^{-1}(U) = f^{-1}(U)$ which is open in $A$ as $f$ homeomorphic.   $g^{-1}$ continuous   Let $U \subseteq A$ be an open set in $A$.  $g(U) = f(U)$ which is open in $B$ as $f$ homeomorphic.  So $g$ is homeomorphic.  Is my understanding correct, this seems almost too straightforward so I was wondering have I overlooked something? Weighting features prior to SVM How do I shrink the partition of an img file made from dd? How does a wizard or bard learn new spells? Show that $S$ is compact. Let $A \in M_n(\bf C)$ and let   $S=\{ A \in M_n(\bf C) : A=A ^*$ and  $\rho (A) \leq 1 \}$, $\rho(A)$ denotes spectral radius of $A$.     Show that $S$ is compact subset of $M_n(\bf C) \cong \bf C^{n^2} $ with usual topology.    $S$ is subset of $\bf O_n(C)$ which is compact so I think its enough to show that $S$ is closed.Please help.