I need something like proof environment from amsthm, but with word "důkaz" (proof in my language) at start of environment. How can I define such environment?  Thanks for help.
123311
I want to change the word "Proof" into "Solution". Of course the following works.  \begin{proof}[Solution] \end{proof}   But it's quite cumbersome to add this additional [Solution] every time. Is there a way to make it automatic?  Edit: I'm using amsthm.  \documentclass{amsbook} \usepackage{amsthm} \begin{document} \begin{proof}[Solution] This is a solution. \end{proof} \end{document}
95811
LyX is great. I use it to write everyday documents. But it seems that its environment can only have one type; different types of environments can't be mixed. For example, you can't have itemize inside a proof. If you try, either you lose the itemize, or the itemize part become outside of the proof. Can I have itemize inside proofs? You need this when making case distinctions.
3707
I use the following custom environment emailHeader and  (credits belong to ).  \definecolor{emailHeaderBackground}{RGB}{225,225,225} \usepackage[framemethod=tikz]{mdframed} \newmdenv[   skipabove=0em,   skipbelow=1em,   hidealllines=true,   backgroundcolor=emailHeaderBackground,   roundcorner=3pt]   {emailHeader}   In the document ...  \begin{emailHeader}   \emailFrom[John Doe]{john.doe@stackexchange.com}   \emailSubject{Re: Please do break apart}   \emailDate{Wed, 22 Aug 2012 12:23:18 +0200}   \emailTo[Jane Doe]{jone.doe@stackexchange.com}   \begin{emailsCc}     \email[Barack Obama]{barack.obama@whitehouse.gov}     \email[Michelle Robinson]{michelle.robinson@whitehouse.gov}   \end{emailsCc} \end{emailHeader}   The output looks like this ...    I use the environment several hundred times in the document. Sometimes it happens that the email header is at the end or beginning of a page. Then, it breaks into separate lines. As you can see the environment is not a verbatim environment for which I found settings to avoid the page break.    Question:  How can I prevent a pagebreak for the above environment? I want the command to be added to the environment definition. I do not want to edit all the places where I use the set of commands.
93654
I have a Predicate Logic argument I need to translate into the symbolism of predicate logic first and then I need to construct a proof in CP.    The argument is "Some wars are just. No war of aggression is just. Therefore, there are wars that are not wars of aggression. (Wx = x is a war, Jx = x is just, Ax = x is a war of aggression)."    I think I have it translated right, however, I am lost on how to construct the proof into CP.    This is how I have translated it (∃x) (Wx • Jx), (x) (Ax → ~Jx) ∴ (∃x) (Wx • ~Ax)   I need help with the CP proof. Any help would be very much appreciated!  I am also only allowed to use implicational rules like, MP, MT, HS, DS, CD, Simp, Conj, and Add.  Equivalence rules such as DN, Com, As, DeM, Cont, Dist, Ex, Re, ME, MI. As well as QN, UI, EG, EI, UG.
78492
I would like to make cross references to own defined counter on environments. So far I have tried the following,   \documentclass{book} \newcounter{entryc}[subsection] \newcommand{\thetheorem}{\thesubsection.\arabic{entryc}} \newenvironment{qd}[1]{\refstepcounter{entryc}\noindent\textbf{Quasi Definition \thetheorem.}\begin{itshape}} {\end{itshape}} % \begin{document}  \chapter{First chapter} This is the first chapter.  \section{First section} This is the first section.  \subsection{First subsection} This is the first subsection.  \begin{qd}{}\label{qd:emp} This is my environment. \end{qd}  \noindent In quasi definition \ref{qd:emp} it \dots  \end{document}   What happens is that     But I would prefer "...In quasi definition 1.1.1.1". Do anyone have some suggestions of how solve this? and How to avoid the          \noindent  thing before the continuation of the text?   Additional information: if possible I would try to not use ntheorem.
54558
Someone please help me! I have some custom environments and I want to label them according to theorem counter. Means, that I get a numbering like Def. 1.1, Thm. 1.2 and so on. Right now, I have to enter the numbers for the labels myself and with growing content of my document, this gets messy really quick.  My goal is a modified environment for definition and theorem such that it automatically includes a label with the name of the current environment (definition, theorem, …) the chapter number and the theorem counter. See my example below.  Does anyone have an idea on how to accomplish this or even a code snippet with an implementation? I am very grateful for any help. :)  Best regards, Joe  \documentclass[a4paper,12pt]{report} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath,amsthm,amssymb,amsfonts,hyperref} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[chapter] \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \begin{document} \chapter{A first chapter} \section{Terminology} \begin{definition}\label{defi:1.1} Some first definition. \end{definition} \begin{theorem}\label{theo:1.2} Some theorem. \end{theorem} Definition~\ref{defi:1.1} is important for Theorem~\ref{theo:1.2}! \end{document}
106959
There are too many options for "Solution/Answer" environments which we can use and I am adopting  for my "big" project that I am doing.  At the moment, it looks like the easiest one for me to implement, but there is just one function to be explored. How can I show/hide the solution environment, so that I can have a teachers' version and a students' version?  When the solutions are hidden, preferably that the space I used for these content are "reserved."  Thanks.
41969
I'm creating a style file, and want to store the body text of the abstract environment so that I can put it on my custom titlepage. I could have required that my users enter the abstract as  \customabstract{Abstract text goes here}`   which I would have implemented as something like  \newcommand{\customabstract}[1]{\gdef\abstracttext{#1}}   However, I would prefer that the user is able to use environment notation,  \begin{customabstract}Abstract text goes here\end{customabstract}   To do this, I thought I could define something like  \newenvironment{customabstract}{\gdef\abstracttext\bgroup}{\egroup}   But this definition makes \abstracttext expand to \bgroup, which is not what I wanted. I tried to solve the problem by \expandafter,  \newenvironment{customabstract}{\expandafter\gdef\expandafter\abstracttext\bgroup}{\egroup}   but this results in a runaway error.  Is there any way to acheive what I want? Answers that helps me understand the inner workings of TeX (instead of just referring to a custom package) are preferred.
22130
I've created my own environment example, similar to theorem. Now I want to generate a list of examples, similar to list of tables or list of figures. How can I achieve this?
123648
It's all in the question really. I am working on a proof in Fitch for a class, but I am very much stuck.  I am proving the tautology that "(P → Q) ↔ (¬P ∨ Q)", and I have already finished half of it, but now I must prove that "(¬P ∨ Q)" implies "(P → Q)". I can't seem to get anywhere.   I try to set up a proof by cases where I assume in different subproofs "¬P" and (in the other) "Q", but then I must prove "P → Q" from those. It seems even more difficult. Any help would be appreciated.
55572
Motivation: If operation is change, and there is no change, then there is no operation. Then the meaning of  1+0 =1 is not in the equality itself but in the comparison with other equalities as  1+2=3. The following definition does not use the meaningless equality  1+0 =1, or generally a*e=a. I guess that trying to prove statements using the next definition, matters/problems will arise that do not arise with the standard definition. Let be the non empty set G = {a,b,c,d,.....z} with the operation *, satisfy the following 3 conditions:  a(bc) = (ab)c, for every a, b, c ∈ G (associative law). For every a,b there is c (unique) such that ac = b. (c  is the &quot;path&quot; from a to b). For every a,b there is d (unique) such that bd = a. (d is the &quot;path&quot; back from b to a).  P.S.: the meaning of 2) and 3) is the possibility to &quot;go&quot;/&quot;move&quot; from any member to any member, through the same path (if a-&gt;b-&gt;c then c-&gt;b-&gt;a as in the Klein group) or through other path (if a-&gt;b-&gt;c then c-&gt;X-&gt;a). Prove: G,*  is a group. Proof: Associativity: From 1) Identity element: Let's prove for every a there is e such that ae=ea=a. Proof: According to 2) For every a,a there is e such that ae = a. (3) According to 2) for every e,a there is x such that ex=a, that means ae =ex =a (4) Let's proof x=a. Multiply by a on the left, such that the result is some g ∈ G: ae = ex = a (4) &lt;---&gt; (a)ae = (a)ex = aa =g &lt;---&gt; aae = (ae)x = aa =g (associative) &lt;---&gt; &lt;---&gt; aae = ax = aa =g (3). (5) According to 2) for every a,g there is x (unique) such that aa=g and ax=g , then a and x are the same, and it's proved x=a. Then replacing in (5): ae = ea = a. Inverse element: Let's prove that for every a, there is b such that a ⋅ b = e and b ⋅ a = e According to 2) for every a,e there is b such that ab = e. (6) According to 2) for every b,e there is X such that bX=e. (7) Let's proof that X=a ab = bX = e (6),(7) Multiply be a on the right: aba =ea a(ba) = ea; Associativity ea = a; Identity element a(ba) = a; then ba = e bX=e according to (7) ba = bX = e; according to 2) for every b,e there is x unique such that bX = e then x=a and it's proved that for every a, there is b such that a ⋅ b = e and b ⋅ a = e.
36135
In the Quantum Operations section in Nielsen and Chuang, (page 358 in the 2002 edition), they have the following equation: $$\mathcal E(\rho) = \mathrm{Tr}_{env} [U(\rho \otimes \rho_{env})U^\dagger]$$ They show an example with $\rho_{env} = |0\rangle \langle0|$ and $U = \mathrm{CNOT}$, and claim that the final solution is: $$P_0\rho P_0 + P_1\rho P_1,$$ where $P_0=|0\rangle \langle0|$ and $P_1=|1\rangle \langle 1|$. These are my steps so far to get this, but I don't know how to trace out the environment after this: Let $\rho$ be $|\psi \rangle \langle \psi |$, so that $\rho \otimes \rho_{env} = |\psi, 0\rangle \langle \psi, 0|$. Applying the unitary $U$, we have $$ |00 \rangle \langle 00| \psi, 0 \rangle \langle \psi, 0 | 00 \rangle \langle 00 |  + |00 \rangle \langle 00| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 | \\  + |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 00 \rangle \langle 00 |  +  |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 |. $$ I don't know how to trace out the environment in the above state. Also, I realize that I have considered only a pure state, if anyone can show it for a general state that would be great.
80994
What is the etiquette to denote the end of an example or remark? For proofs, it's the QED symbol (white box). I know it's not necessary for everyone, but I would like to use it to break up the text a little. What should I use? A circle or filled box or what?
117623
I would like to display \begin{verbatim}...\end{verbatim} inside the environment verbatim like in the folowing piece of code. Is it possible ?  \begin{verbatim} Bla, bla, \begin{verbatim}...\end{verbatim} Bla, bla,... \end{verbatim}
112991
I came across the following proof in the book Logic, by Paul Tomassi:   (P & Q) → ~R : R → (P → ~Q)   According to the author, the proof should be a simple application of modus tollens. The following is perhaps the most obvious way of proving it:    1.   (P & Q) → ~R         Premise  2.   R                    Assumption  3.   ~~R                  Double Negative Ins.  4.   ~(P & Q)             1,3 Modus Tollens  5.   ~P ∨ ~Q              4 DeMorgen  6.   P → ~Q               5 Implication Ident.  7.   R → (P → ~Q)         2,6 Conditional Proof   The only problem is that the author hasn't yet introduced DeMorgen's identity or any other identities. The following is a list of all the resources that have so far been introduced:   Conjunction Introduction &amp; Elimination Conditional Introduction &amp; Elimination (Modus Ponens) Biconditional Introduction &amp; Elimination Double Negative Introduction &amp; Elimination Modus Tollens   However, I don't see any way to do the proof without identity rules. The following is a possible strategy for trying to solve it without identies:   1.   (P & Q) → ~R         Premise 2.   R                    Assumption 3.   ~~R                  Double Negative Ins. 4.   ~(P & Q)             1,3 Modus Tollens 5.   ~(P → ~Q)            Assumption ... 15.  ~(P → ~Q) → (P & Q)  5,14 Conditional Proof 16.  P → ~Q               4,15 Modus Tollens 17.  R → (P → ~Q)         2,16 Conditional Proof   But I don't see how it's possible to get to my theoretical step 15 without deriving the identies from scratch. According to the author, the proof is possible in 11 steps.  How can it be done?
112162
Is there a way to create an environment for reactions defined with mhchem  that behaves similarly to equation in terms of numbering the reactions (including a \nonumber command)? Can I get a list of those reactions?  \documentclass{article} \usepackage[version=3]{mhchem}  \newcounter{rxnum}  \newenvironment{reaction}{ \refstepcounter{rxnum} \noindent %??? }{ %??? \par\noindent }  \begin{document}  \begin{reaction}  \ce{X + Y -&gt; XY} \end{reaction}  \begin{reaction}  \ce{XX + YY -&gt; X2Y2} \\%&lt;- newline just like in equation  \ce{A + B -&gt; C} \end{reaction}     \end{document}   This is a two part question. The first part is about creating an environment that behaves similarly to equation, but for reactions defined in mhchem. I had a look in source2e in the definition of equation (around l. 293 in source2e) but couldn't quite figure out how to transfer that onto my new environment.  The second part is about how to create a list of those reactions containing the reaction number, the page they occur on and optional a description that is only printed in the \listofreactions.  For example:  In the code I write  \begin{reaction} \ce{X + Y -&gt;XY}\descriptionforlist{XY formation reaction}\\ \ce{A +B -&gt; C} \ce{G + H -&gt; J}\nonumber     \end{reaction}   Then in the list it should appear as  List of reaction  R 1: XY formation reaction                6  R 2:                                      6
11452
I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).  I am using the notaion of David Marker's book "Model theory: an introduction".