I've seen $\mathsf{ZFC2}$ mentioned in a few questions  and I'm curious about ways to axiomatize it that have the fewest differences from plain $\mathsf{ZFC}$ as possible. My question is similar to , but proposes an explicit theory and is specifically about minimizing the differences from first-order $\mathsf{ZFC}$.  mentions that $\mathsf{ZFC2}$ has as axioms the second-order versions of $\mathsf{ZFC}$ axioms, but doesn't go into much detail about what exactly that means. For the purposes of this question, I'm assuming that there's rough consensus on what $\mathsf{ZFC2}$ really is as a theory and am wondering if my naive construction is equivalent to it.  For the sake of concreteness, let the language second-order logic be the same as the language of first-order logic but with second order quantifiers $\forall x : R(n)\mathop. \cdots$ and $\exists x : R(n) \mathop. \cdots$, which quantify over relations of arity $n$. A bare quantifier $\forall x \mathop. \cdots$ is a first-order quantifier. Does naively changing the axiom schema of $\mathsf{ZFC}$ so that $\varphi$ ranges over all second-order formulas give an equivalent theory to $\mathsf{ZFC2}$? $\mathsf{ZFC2}$ has two axiom schemas:  and . If we treat $\forall x \mathop. \cdots$ and $\exists x \mathop. \cdots$ as implicitly first-order quantification, we can trivially transfer all the non-schematic axioms of $\mathsf{ZFC}$ to a second order setting. If we then take another look at the axiom schema of specification and the axiom schema of replacement and allow the synctactic variable $\varphi$ to range over all the second-order sentences, do we get an axiomatization of $\mathsf{ZFC2}$ or a theory equivalent to $\mathsf{ZFC2}$?
Below, the  of second-order logic is used.    My question is about a second-order analogue of $ZFC$ other than the usual "second-order $ZFC$." Rather than define the latter, I'll just state that its (set-sized) models are exactly the $V_\kappa$s with $\kappa$ strongly inaccessible (or if you prefer, the uncountable Grothendieck universes).  Specifically, I'm interested in the second-order theory consisting of the usual (first-order) axioms of Pairing, Extensionality, Union, Choice, Infinity, and Powerset, and the Separation and Replacement schemes ranging over second-order formulas. I'll call this "$ZFC_2^{scheme}$."    Specifically, for each formula $\varphi(x, y_1,...,y_k)$ of second-order logic with only the displayed variables, each of which is first-order, we have a corresponding Separation and Replacement instance. We do not allow second-order parameters here, so formulas with second-order free variables don't yield Separation/Replacement instances.   The passage from $ZFC$ to $ZFC_2^{scheme}$ has a nice uniformity: it's an example of a more general operation $\mathcal{ZFC}$ which spits out a $ZFC$-analogue given a logic at least as strong as first-order logic.    This is the reason for the parameter restriction above: I want $ZFC_2^{scheme}$ to look just like $\mathcal{ZFC}(\mathcal{L})$ for any other logic $\mathcal{L}$, and not take into account the particular feature that we now have a new kind of variable at our disposal - and in particular, take as "syntax-free" an approach to logics as we can get away with. (This is a choice of arguable value, but it's the one I'm making for now.)   It's easy to show that every model of $ZFC_2^{scheme}$ is well-founded, so to understand its models we only need to look at transitive sets. A natural hope at this point is that $ZFC_2^{scheme}$ is just second-order $ZFC$ in disguise, that is, that for a transitive set $M$ we have $M\models ZFC_2^{scheme}$ iff $M=V_\kappa$ for some strongly inaccessible $\kappa$. However, this isn't at all obvious to me (although it is easy to show the right-to-left direction).     Question. What are the (set-sized) models of $ZFC_2^{scheme}$?   (Note by contrast that the "arithmetic analogues" $PA_2^{scheme}$ and second-order $PA$ are equivalent since each pins down $\mathbb{N}$ up to isomorphism - although that leads to .)  (I'm happy to drop Choice if that would help.)  EDIT: As Hanul Jeon pointed out below,  shows that consistently $ZFC_2^{scheme}$ (there called "$ZFC_2^{def}$") has countable models, so we have a partial answer. However, I don't see at the moment a way to get an outright $ZFC$ proof that $ZFC_2^{scheme}$ is strictly weaker than second-order $ZFC$.
I have a dll that was written in c++, I need to use this dll in my c# code. After searching I found that using P/Invoke would give me access to the function I need, but these functions are defined with in a class and use non-static private member variables. So I need to be able to create an instance of this class to properly use the functions. How can I gain access to this class so that I can create an instance? I have been unable to find a way to do this.   I guess I should note that the c++ dll is not my code. Reload the page on hitting back button Lets say I have a loop in Bash:  for foo in `some-command` do    do-something $foo done   do-something is cpu bound and I have a nice shiny 4 core processor. I'd like to be able to run up to 4 do-something's at once.  The naive approach seems to be:  for foo in `some-command` do    do-something $foo &amp; done   This will run all do-somethings at once, but there are a couple downsides, mainly that do-something may also have some significant I/O which performing all at once might slow down a bit. The other problem is that this code block returns immediately, so no way to do other work when all the do-somethings are finished.  How would you write this loop so there are always X do-somethings running at once? When does a dual of a compound proposition equal itself? So I am studying computer science and right now I am stuck on a problem.  When does s∗ = s, where s is a compound proposition?  So far the only thing I can come up with is:  s* = s when the compound proposition is composed only of the same           propositions. (ex. p ∧ p = p ∨ p)  The book defines duality as:  The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Discrete Mathematics and its Applications, Rosen, 7e)  Any help would be great, this is a tricky one. This started to happen immediately after I had rebooted the first time after doing a system upgrade. It first starts with a dialogue that says "System program problem detected". Then when I try to hit 'report problem' not much happens. I am led through a dialogue that always ends up the problem cannot be solved.  I am aware this is not a lot of information, however I'm not sure which information I need to publish and how should I obtain it to debug this problem.  Here's a screenshot! There's this exercise: let $\,f\,$ be analytic on $$D:=\{z\;\;;\;\;|z|&lt;1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D$$  and $\,z=0\,$  a zero of order $\,m\,$ of $\,f\,$.  Prove that $$\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m$$  My solution: Induction on $\,m\,$: for $\,m=1\,$ this is exactly the lemma of Schwarz, thus we can assume truth for $\,k&lt;m\,$ and prove for $\,k=m&gt;1\,$ . Since $\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,$ analytic in $\,D\,$ ,  put  $$g(z):=\frac{f(z)}{z}=z^{m-1}h(z)$$   Applying the inductive hypothesis and using Schwarz lemma $\,\,(***)\,\,$ we get that  $$|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|$$ and we're done...almost: we still have to prove $\,|g(z)|\leq 1\,$ for all $\,z\in D$ in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with $\,z\in D\,$):  $(1)\,\,$ For $\,f(z)=z^mh(z)\,$ we apply directly Schwarz lemma and get $$|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1$$ And since now the function $\,f_1(z)=z^{m-1}h(z)\,$ fulfills the conditions of S.L. we get  $(2)\,\,$ Applying again the lemma,  $$|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1$$and now the function $\,f_2(z):=z^{m-2}h(z)\,$ fulfills the conditions of them  lemma so...etc.  In the step$\,m-1\,$ we get  $$|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,$$  and this is what allows us to use the inductive hypothesis in $\,\,(***)\,\,$ above.  My question: Is there any way I can't see right now to deduce directly, or in a shorter way, that $\,|h(z)\leq 1\,$ ? Short Story about interminable rain on Venus It's not All Summer in a Day.  This was a story about explorers on Venus, who are separated from their base (possibly they're lost, possibly they suffered equipment failure, possibly they're just traveling from base to base) and are slowly being driven insane because it never stops raining.  The story is old enough that Venus' clouds were still believed to indicate a wet climate.  I can't remember much more detail than that.  I have the impression that the narrator at one point either dreamed of or fantasized about walking into the destination base and being handed a very plush towel to dry off.  I can't remember the central plot points - only the (very wet) setting.  I would have read this in the late 1970's or early 1980's.  I thought it might be an H. Beam Piper short story but I've been through his catalog and don't see any likely suspects. White screen of death: Fatal error: Allowed memory size of X bytes exhausted I have a issue with my Drupal installation, for example:   when I enable or disable the modules, It redirects me to a blank page, when I create a new content type and save it, redirects me to a blank page, when I add a new view and save it, it redirects me to a blank page, when I try to clear the cache, it redirects me back to a blank page, or in similar cases.   Basically all the confirmation pages redirect me to a white screen. When I refresh it again, it shows me the page. I tried to increase the PHP memory value but it doesn't help.  Are there any other solutions for this?  The error which I'm having:      Fatal error: Allowed memory size of 100663296 bytes exhausted (tried to allocate 8192 bytes) in sites/all/modules/views/plugins/views_plugin_localization_none.inc on line 1 I am really satisfied that $(x \to a) ≠(x=a)$ and if that is not right , Then all the process of $Limits$ is dividing by zero and that is a crime.  Since $(x \to a) + h = (x=a)$ , $h ≠ 0$,So Why does $f(x \to a) = f(x = a)$ ?  NOTE:I am talking about continuous function. Continuous function preserving rational difference. Find all the continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that if $x-y\in \mathbb{Q}$ then $f(x)-f(y)\in \mathbb{Q}$.   Can somebody give me some hint to solve this problem. I think that the functions are $f(x)=mx+b$ where $m\in\mathbb{Q}$.  Thanks! Git SSH error: "Connect to host: Bad file number" What's the difference between an Application, a Process, and a Service? What is the best way to replace the file browse button in html? I know that it's possible to replace the browse button, which is generated in html, when you use input tag with type="file.  I'm not sure what is the best way, so if someone has experience with this please contribute. What is the difference between Python's list methods append and extend? The classical book on commutative algebra Introduction to Commutative Algebra, by Atiyah and Macdonald, has the following as exercise I.11.       A ring is Boolean if $x^2=x$ for any $x$ of $A$. In a Boolean ring $A$, show that     i) $2x=0$ for all $x\in A$;    ii) Every prime ideal of $A$ is maximal, and its residue field consists of two elements;   iii) Every finitely generated ideal of $A$ is principal.     I can but solve the first two of them; for the last one, I cannot even perceive the situation here, for it will not, generally speaking, be a Dedekind domain, whereof my knowledge is somewhat more. I have hitherto tried to find one element for each ideal of two generators which can stand the stead of them, and then proceed by induction. But even the first step fails to produce any result. Thanks for any hints and inspirations in advance. How can I search for bounty questions? Is there any possible way to search and list the bountied questions separately in SE? Is there a JavaScript equivalent of Java's class.getName()? Is 225.225.225.128 a valid subnet mask? Python is quite cool, but unfortunately, its debugger is not as good as perl -d.    One thing that I do very commonly when experimenting with code is to call a function from within the debugger, and step into that function, like so:  # NOTE THAT THIS PROGRAM EXITS IMMEDIATELY WITHOUT CALLING FOO() ~&gt; cat -n /tmp/show_perl.pl 1  #!/usr/local/bin/perl 2 3  sub foo { 4      print "hi\n"; 5      print "bye\n"; 6  } 7 8  exit 0;  ~&gt; perl -d /tmp/show_perl.pl  Loading DB routines from perl5db.pl version 1.28 Editor support available.  Enter h or `h h' for help, or `man perldebug' for more help.  main::(/tmp/show_perl.pl:8):    exit 0;  # MAGIC HAPPENS HERE -- I AM STEPPING INTO A FUNCTION THAT I AM CALLING INTERACTIVELY   DB&lt;1&gt; s foo()  main::((eval 6)[/usr/local/lib/perl5/5.8.6/perl5db.pl:628]:3): 3:      foo();     DB&lt;&lt;2&gt;&gt; s main::foo(/tmp/show_perl.pl:4):     print "hi\n";     DB&lt;&lt;2&gt;&gt; n hi main::foo(/tmp/show_perl.pl:5):     print "bye\n";     DB&lt;&lt;2&gt;&gt; n bye     DB&lt;2&gt; n Debugged program terminated.  Use q to quit or R to restart,   use O inhibit_exit to avoid stopping after program termination,   h q, h R or h O to get additional info.      DB&lt;2&gt; q   This is incredibly useful when trying to step through a function's handling of various different inputs to figure out why it fails.  However, it does not seem to work in either pdb or pydb (I'd show an equivalent python example to the one above but it results in a large exception stack dump).  So my question is twofold:   Am I missing something? Is there a python debugger that would indeed let me do this?   Obviously I could put the calls in the code myself, but I love working interactively, eg. not having to start from scratch when I want to try calling with a slightly different set of arguments. Syntax error due to using a reserved word as a table or column name in MySQL