I want to prove $\alpha \rightarrow \beta, (\alpha \rightarrow \lambda) \rightarrow \beta $ proves $\beta$ in HPC  I use MP and the axioms listed here:   I guess that the end of the proof should look something like this:  $(\alpha\rightarrow(\beta\rightarrow\lambda)) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow\lambda))$  $(\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow\lambda)$   $\alpha \rightarrow \lambda$  $\beta$  but I can't find out how I can get to $(\alpha\rightarrow(\beta\rightarrow\lambda))$ given my premises  would be happy for some help
I'd really like your help proving:  $(A\rightarrow B),(A\rightarrow C)\rightarrow B,  \mapsto_{HPC} B $  Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:   $A\rightarrow(B \rightarrow A)$ $(A\rightarrow(B\rightarrow C)) \to ((A\rightarrow B)\rightarrow(A\rightarrow C))$ $(A\rightarrow B)\rightarrow ((A\rightarrow\bar{B})\rightarrow \bar{A})$ $\bar{\bar{A}} \rightarrow A$   In addition tried to use these following lemmas: $\bar{A} \rightarrow (A \rightarrow C) $ and $(A\rightarrow B)\rightarrow (\bar{B} \rightarrow \bar{A})$.  Any suggestions?
What does “would not have been possible had I remained” mean? What does this mean?      I am sure this would not have been possible had I remained a typical Anglophone North American.   I have never heard that before and I really find it strange.  I deduce that it means      I am sure this would not have been possible if I kept remaining Anglophone North American.   But I might be wrong in this. Analogy of Pearson correlation for 3 variables 0:root@SERVER:/root # echo "something" | sed -e 's/^/\t/g' tsomething 0:root@SERVER:/root #    AIX/ksh .. why doesn't it works? I just want to put a "tab" before every line! Duality discrete math problem This is the only answer I got wrong on my HW and the prof does not want to give us the correct answers before our midterm     The dual of a compound proposition that contains only the logical operators $\lor$ , $\land$ , and   $\neg$ is the compound proposition obtained by replacing each $\lor$ by $\land$ , each $\land$ by $\lor$ , each $\def\T{{\rm T}}\def\F{{\rm F}}$   $\T$ by $\F$ , and each $\F$ by $\T$ . The dual of $s$ is denoted by $s^*$. Find the dual of these compound propositions.      a) $p \lor\neg q$     I got $\neg p \land q$       b) $p \land (q \lor (r \land \T))$     My answer was $\neg p \lor (\neg q \land r)$       c) $(p \land \neg q) \lor (q \land \F)$     My answer was $(\neg p \lor q) \land  \neg q$  I have tried googling the problem and cannot come up with anything on duals and our lectures are online and upon reviewing do not see anything. I am just confused and looking for a little guidance on what was incorrect with my answers. I am about to sell my old desktop PC and I am cautious about some of my sensitive information being available to the purchaser, even after reformatting the hard-drive, using data recovery software.  How can I securely wipe the hard drive so that the data on it cannot be recovered?  Although I specifically want help with my Windows PC, it wouldn't hurt if there were suggestions for Macs as well. Are console games locked to a certain region? I'm going on a vacation from Europe to the US and was planning of bringing along some Xbox360 or PS3 games, but now I'm curious if they will be compatible with my European Xbox.  In that past there would be a clear difference between NTSC and PAL games, but is this still the case with the current gen of consoles? I am trying to define the probability distribution of $Z$ such as $Z = X_1\cdot X_2$ where $X_1$ and $X_2$ are two independent and identically exponentially distributed variables.  $$P(X_1=x) = \lambda e^{-\lambda x}$$ $$P(X_2=x) = \lambda e^{-\lambda x}$$  I tried something like that….  $$P(Z=x) = P(X_1 = a) \cdot P(X_2 = x/a)$$  Is this first equation correct?  $$P(Z=x) = \lambda e^{-\lambda a} \cdot \lambda e^{-\lambda x/a} = \lambda^2 e^{-\lambda (a+x/a)}$$  I doesn't feel like I found the solution as the dummy variable $a$ remains in the final result. Can you please help me solving this problem? Can I ignore coefficients for non-significant levels of factors in a linear model? After seeking clarification about linear model coefficients  I have a follow up question concerning non-signficant (high p value) for coefficients of factor levels.  Example: If my linear model includes a factor with 10 levels, and only 3 of those levels have significant p values associated with them, when using the model to predict Y can I choose to not include the coefficient term if the subject falls in one of the non-signficant level?  More drastically, would it be wrong to lump the 7 non-significant levels into one level and re-analyze? Is there a replacement for unistd.h for Windows (Visual C)? I'm porting a relatively simple console program written for Unix to the Windows platform (). All the source files include "unistd.h", which doesn't exist.  Removing it, I get complaints about misssing prototypes for 'srandom', 'random', and 'getopt'. I know I can replace the random functions, and I'm pretty sure I can find/hack-up a getopt implementation.    But I'm sure others have run into the same challenge.  My question is: is there a port of "unistd.h" to Windows? At least one containg those functions which do have a native Windows implementation - I don't need pipes or forking.  EDIT:  I know I can create my very own "unistd.h" which contains replacements for the things I need - especially in this case, since it is a limited set. But since it seems like a common problem, I was wondering if someone had done the work already for a bigger subset of the functionality.  Switching to a different compiler or environment isn't possible at work - I'm stuck with Visual Studio. What is the meaning of the beta-coefficient for an interaction term in a crossover study? I have asked a similar question here:  I am puzzled by the interpretation for an interaction term. In my data my Y is an interval variable with the health outcome of an experiment. I have used an interaction term in which I have interacted the condition with the predisposition of the subject considering health status at base level. They are both categorical variables (factor variables in R).  Now it gets complicated because the Condition was two treatments: in treatment A subjects got the placebo first and then the real medicine whereas in treatment B they go the real medicine first and the placebo second. All it changes is the order.   Health outcome = a + Condition * Health.Base   I have the worst state of health at the base level as my reference category I find that interaction with the Condition is statistically significant but I am not sure how to interpret this.  I use the lm() function of R (although my design looks more like an ANOVA) and in the output I get the b coefficient in an output that looks like this:  ConditionB:Health.Base.So.and.So         (Beta and p-value) ConditionB:Health.Base.Excellent         (Beta and p-value)   A statistically significant interaction term for those in Excellent health at baseline would mean that they are affected by the Condition B more than Condition A compared to the reference category people (Poor health at baseline). Is this right? What does the beta-coefficient represent?   If I would like to examine for each Health category at the base line separately without comparing to a reference category I would have to code each category as a dummy variable. However, in this case I would compare whether membership to a specific health status at the base line significantly changes between conditions compared to those who belong to the other health statuses at the base line. Is this right? Again, what does the beta-coefficient represent?  Would it be right to assume that the choice of the interaction between the Condition and the dummy variables is easier to interpret?  --- EDIT --- R output:  Call: lm(formula = HealthOutcome ~ Condition * HealthStatus,      data = datA)  Residuals:     Min      1Q  Median      3Q     Max  -1.5957 -0.5942 -0.2640  0.4423  2.4423   Coefficients:                                        Estimate Std. Error t value     Pr(&gt;|t|)     (Intercept)                            2.595652   0.053044  48.934  &lt; 2e-16 *** ConditionCondB                        -0.001449   0.077071  -0.019    0.985     HealthStatusSo.and.So                 -0.331693   0.078094  -4.247  2.35e-05 *** HealthStatusExcellent                 -0.836724   0.092692  -9.027  &lt; 2e-16 *** ConditionCondB:HealthStatusSo.and.So   0.137490   0.110612   1.243    0.214     ConditionCondB:HealthStatusExcellent  -0.199787   0.133943  -1.492    0.136     --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 0.8045 on 1059 degrees of freedom   (68 observations deleted due to missingness) Multiple R-squared:   0.16, Adjusted R-squared:  0.156  F-statistic: 40.33 on 5 and 1059 DF,  p-value: &lt; 2.2e-16 How to encrypt folders (not in Home)? ASP.NET -> WCF concurrency problem If you draw two cards, what is the probability that the second card is a queen? We had this question arise in class today and I still don't understand the answer given. We were to assume that drawing cards are independent events. We were asked what the probability that the second card drawn is a queen if we take two from the deck. The answer given was 4/52, which seems counter-intuitive to me. How is the probability still 4/52 if there was a card drawn before it? What if the first card drawn was a queen? The -k option (or --convert-link) will convert links in your web pages to relative after the download finishes, such as the man page says:     After the download is complete,   convert the links in the document to   make them suitable for local viewing.   This affects not only the visible   hyperlinks, but any part of the   document that links to external   content, such as embedded images,   links to style sheets, hyperlinks to   non-HTML content, etc.   So, if I didn't specify -k, can I run wget again after the download and fix that, and if so, what would be the proper command? My guess is wget -c [previous options used] [url] and run it in the same working directory as the file were downloaded to. Tikzpicture-figure: Order plot data (asc/desc)? I am using a tikzpicture for plotting. the values I want to plot are typical (X,Y) pairs. The x-value is an uninteresting id-value. I want to order the bins ascending/descending according to the y-value, but how do I do that? The figure seems to automatically take the x-values for ordering, how can I override that? (I am not particularly fixed to tikzpicture - its just the first thing that actually worked and displayed the plot....) Following minimal-working example, how could I achieve a reordering according to the y-value?  \documentclass{article} \usepackage[english]{babel} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{relsize} \usepackage{times} \usepackage{url} \usepackage{latexsym} \usepackage{graphicx} \usepackage{colortbl} \usepackage{color} \usepackage{caption} \usepackage{pgfplots, pgfplotstable} \usetikzlibrary{arrows} \usepackage{amsmath} \usepackage{multirow} \usepackage{booktabs}  \usepackage{filecontents} \definecolor{OgAns}{rgb}{0, 0.8, 0.4}  \begin{filecontents}{testdata.dat} 1   30 2   44 4   26 3   39 \end{filecontents}  \begin{document} \begin{figure} \begin{tikzpicture}         \begin{axis}[                ybar stacked,             ymin=0,             ymax=100,             bar width=5pt,             legend style={at={(0.35, -0.4)},anchor=south west},             legend columns=-1                       ]           \addplot[ybar,fill=OgAns]  file {testdata.dat};          \end{axis}     \end{tikzpicture} \end{figure}   \end{document} I have a loop going, but there is the possibility for exceptions to be raised inside the loop. This of course would stop my program all together. To prevent that I catch the exceptions and handle them. But then the rest of the iteration runs even though an exception occurred. Is there a keyword to use in my except: clause to just skip the rest of the current iteration? Need to delete a directory with name '~' created accidentally Question about probable fixed use of "what do you do?" and "how do you do?" Do we always have to use the question how do you do? fixed in greeting people? Can we ask it in other situations rather than greetings? Like when we are wondering how someone is doing the given task.  Do we always have to use the question what do you do? for asking about sojeone's job? Can we use it sonetimes instead of ** what are you doing?** Set of discontinuity of monotone function is countable The result I am trying to prove is the following:     Let $F:\Bbb R\to\Bbb R$ be increasing. Then the set of points, at which it is discontinuous, is countable.   I have been reading this form Folland's Real Analysis and he proves this by considering the sum $\sum_{|x|&lt;N}[F(x^+)-F(x^-)]$ which has to be finite by some sort of telescoping sum. Now, I am not sure how he defines this sum. I am assuming he does this by using nets (let me know if there is another way to look at this), but he doesn't introduce the concept of nets until the next chapter. So I was looking for some other ways to prove this result and I found the following result (from Stein-Shakarchi) which is quite similar but assumes $F$ to be bounded.      A bounded increasing function $F$ on $[a,b]$ has at most countably many discontinuities.   This is shown by using the density of rationals in $\Bbb R$, i.e. $\exists r_x\in\Bbb Q$ s. t. $F(x^-)&lt;r_x&lt;F(x^+)$ corresponding to each point of discontinuity $x$.   Now I feel like there shouldn't be a problem in doing the exact same thing for unbounded increasing function. Is this right? Secondly how does this proof relate to the first one? I feel like there isn't much of a difference between the two proofs. To be precise I think that the previous sum is well defined/finite because of this countability of discontinuous points. What do you think about this way of thinking?  Edit. Here is the Folland's proof in more detail:  Since $F$ is increasing, the intervals $(F(x^-),F(x^+))$ are disjoint for all $\forall x\in\Bbb R$. Moreover,  $\forall|x| &lt; N$ such intervals are contained in $(F(-N),F(N^+))$ and so  $$\sum_{|x|&lt;N}[F(x^+)-F(x^-)] \leqslant F(N) - F(-N) &lt; \infty. $$ Hence the set $\{ x \in (-N,N) : F(x^+) \neq F(x^-)\}$ is countable. I can't figure out how to turn on the Bluetooth on my Nikon D5600. I have updated the firmware to the latest 1.01 and still the same problem.   Bluetooth is dimmed out on the camera and will not turn on. I get an error message saying, "This option is not available at current settings or in the camera's current state".  I have tried everything I can think of and no go. I can't find help for this problem anywhere on the internet and still no reply from Nikon.  How do I turn on my camera's Bluetooth?