I'm trying to typeset some basic logic proofs in LaTeX. I need a multiple column layout, and I need the lines numbered on the right. I want to reset the numbering for each proof. The output should look like this:
1) (x)¬F A
2) (Ex)F
3) ...
What is the easiest way to do this? enumerate doesn't seem a good idea, since it's not going to like the multiple column things. I need everything to be in mathmode as well.
Is there an easy way to do this?
[edit]
Here is what I've managed to do so far, but it is far from elegant.
\begin{minipage}{\linewidth}
\boldmath \center{$ \forall x \neg F \to \neg \exists x F $} \unboldmath
\begin{math}
\begin{array}{llr}
1. &\forall x \neg F & \\
2. &\exists x F & A \\
3. &F[\alpha / x ] & ES 2 \\
4. &\neg F[\alpha / x] & US \\
5. &\exists x F \to F[\alpha / x ] & CP 2,4 \\
6. &\neg \exists x F & TI 4,5
\end{array}
\end{math}
\end{minipage}
Ideally I want something that automatically numbers steps and makes the "title" line easier...
See the packages for Fitch-style proofs here. You simply don't need any subproofs. LaTeX for Logicians also covers other proof styles, logical symbols, etc.
\newcount\mycount
\def\nextnum{\global\advance\mycount 1 \the\mycount.}
\def\logicx#1{\halign{$\nextnum##$\quad \hfil&$##$\hfil
&\qquad\hfil##\crcr#1\crcr}\endgroup}
\def\logic{\begingroup \global\mycount 0 \let\\\cr \logicx}
\logic{
&\forall x \neg F & \\
&\exists x F & A \\
&F[\alpha / x ] & ES 2 \\
&\neg F[\alpha / x] & US \\
&\exists x F \to F[\alpha / x ] & CP 2,4 \\
&\neg \exists x F & TI 4,5 \\
}
Check out section 3.6 of the amsmath package documentation.
Sorry, don't have a \LaTeX compiler now, so my code may be not compileable or contain silly mistakes.
You need a new counter for your lines:
\newcount\logicproofline
Then, you need a command to start it:
\def\proofline#1{
\global\advance\logicproofline by 1
\shoveright{\text{\the\logicproofline}\qquad#1}
}
Now, you would need to redefine proof to reset your counter:
\AtBeginDocument{
\let\myoldproof=\proof
\def\proof{\myoldproof\logicproofline=0\relax}
}
The redefinition is needed because proof is not associated with any counter, so we cannot make that one a master counter.
Related
How to typeset the following numbered mathematical formula:
Use amsmath's cases environment:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\Delta(x) = \begin{cases}
1 & \text{if $x > 0$} \\
-1 & \text{if $x < 0$} \\
0 & \text{otherwise}
\end{cases}
\]
\end{document}
Here is one way to do it:
$
\Delta(x)=
\left\{
\begin{matrix}
1 & \mathrm{if}~x>0 \\
-1 & \mathrm{if}~x<0 \\
0 & \mathrm{otherwise}
\end{matrix}
\right.
$
Which produces the equation you wanted like this:
The example LaTeX equation
This type of equation has a few key tricks in it:
The left bracket has to be declared as \{ in order for LaTeX to
know to display it since it is a special character
Also, the left brace has to be resized to match however many rows you include in your equation, which is done by the \left command preceding it. This can also be done for (, [, etc.
Since you have a \left command, it has to be accompanied by a \right}, but you don't want to display the right brace, so the \right. ends the section without displaying the right brace.
The \mathrm command turns the text inside it back to normal text (removing the math mode temporarily) so that your "if" and "otherwise" aren't italicized.
The ~ just adds a little space so that your "if" doesn't collide with the x > 0, which often happens when using mathrm
Alternatively, you can check out this post for other ways to write piecewise functions in LaTeX. I hope this helps.
I'm using an Align environment inside of a proof, and I'm getting the error "Missing $ inserted." I've commenting out all but the first line of the align makes the error go away, but I still can't figure out what the problem is. No solution I could find works. Am I missing something?
\documentclass[12pt,letterpaper]{article}
\usepackage{ifpdf, enumerate}
\usepackage{mla}
\usepackage{gb4e}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{stmaryrd}
\newcommand{\tuple}[1]{\ensuremath{\left \langle #1 \right \rangle }}
\newcommand{\evaluation}[2][]{\ensuremath{\llbracket #2\rrbracket^{#1}}}
\begin{document}
\begin{proof}
\begin{align*}
\evaluation[\mathbb{M}]{(Mx)(D (x) \& V (x))} = 1 \iff \\
\evaluation[\mathbb{M}]{\lbrack \alpha / x \rbrack (D (x) \& V (x))} = 1 \text{for most constants \alpha} \\
\evaluation[\mathbb{M}]{(D (\alpha) \& V (\alpha))} = 1 \text{for most constants \alpha } \\
\evaluation[\mathbb{M}]{D (\alpha)} = 1 \text{and} \evaluation[\mathbb{M}]{V (\alpha)} = 1 \text{for most constants \alpha } \\
F(\alpha) \in F(D) \text{and} F(\alpha) \in F(V) \text{for most constants \alpha } \\
\intertext{Examining the definitions of the model, we see that there are only four entities that are both in $F(D)$ and $F(V)$: Alice, Bob, Colin, and David. Because this is not more than half of the total twelve entities in the universe, our naive definition of ``most'' is not satisfied. Therefore the statement is false, which implies that the truth value of the origenal proposition is not 1.}
\end{align*}
\end{proof}
\end{document}
Here is what the error message looks like.
./test.tex:24: Missing $ inserted.
<inserted text>
$
l.24 \end{align*}
?
You've missed only one tricky fact: \alpha only works in math mode, so if you replace all occurrences of
\text{for most constants \alpha} \\
(lines 18, 19, 20, 21) with
\text{for most constants }\alpha \\
then your tex will compile sweetly. Also:
\text{for most constants}\ \alpha \\
or:
\text{for most constants $\alpha$} \\
as you already did in line 22 of your code (...$F(D)$ and $F(V)$...).
This answer is not applicable to this exact case, but the given error can also pop up if the "amsmath" package is not being used. It becomes tricky because the whole code looks alright except I had forgotten to add the \usepackage{amsmath} line in the beginning of the document.
I want to create a cycle like this in LaTeX:
http://upload.wikimedia.org/math/0/c/8/0c814b6d3ade0dfd8ea4408af6e07ac6.png
Instead of numbers I want to have b1,b2,...,b9 and m1,m2,...,m9.This is the code I use:
\pi_0 = \binom{b_1\ b_2\ b_3\ b_4\ b_5\ b_6\ b_7\ b_8\ b_9}{m_1\ m_2\ m_3\ m_4\ m_5\ m_6\ m_7\ m_8\ m_9}
But it does not produced the desired result, as the b-s and m-s are not aligned (I want for them to be exactly one above the other):
Could you help me, please?
Try using the array environment to produce a matrix:
\pi_0 = \left(
\begin{array}{cccc}
b_1 & b_2 & b_3 & b_4 \\
m_1 & m_2 & m_3 & m_4
\end{array}
\right)
By the way, there is a separate StackExchange site for LaTeX.
I am trying to align the following equations around the equal sign. What can I do? Thanks! I am using the AMSMath package
$$m\ddot{x}_{1}-K(x_{2}+x_{12})+C(\dot{x}_{12}+\dot{x}_{2})+2C\dot{x}_{1} = fx_{1}$$
$$m\ddot{x}_{2}-K(x_{3}+x_{1})+C(\dot{x}_{1}+\dot{x}_{3})+2C\dot{x}_{2} = fx_{2}$$
$$m\ddot{x}_{3}-K(x_{4}+x_{2})+C(\dot{x}_{2}+\dot{x}_{4})+2C\dot{x}_{3} = fx_{3}$$
\begin{align}
m\ddot{x}_{1}-K(x_{2}+x_{12})+C(\dot{x}_{12}+\dot{x}_{2})+2C\dot{x}_{1} &= fx_{1} \\
m\ddot{x}_{2}-K(x_{3}+x_{1})+C(\dot{x}_{1}+\dot{x}_{3})+2C\dot{x}_{2} &= fx_{2} \\
m\ddot{x}_{3}-K(x_{4}+x_{2})+C(\dot{x}_{2}+\dot{x}_{4})+2C\dot{x}_{3} &= fx_{3}
\end{align}
See the user's guide.
To make a long story short... the following idea of using eqnarray is kind of obvious, but turns out to be a really bad idea. eqnarray brings lots of issues which are solved in amsmath and related packages. Read Lars Madsen: Avoid eqnarray for the details.
\begin{eqnarray*} %% Do avoid eqnarray if possible.
x_1 & = & 1 \\
x_2 & = & 2 \\
x_3 & = & 3
\end{eqnarray*}
(Answer revised with information from Arthur Reutenauer, Thomas and the Madsen paper linked to by Thomas.)
There are several packages to do what you want, one of the most widely used is the AMS' align environment from the amsmath package.
Simplified extract from amsldoc.pdf:
\begin{align}
x&=y \\
x’&=y’ \\
x+x’&=y+y’
\end{align}
The equations are aligned around the symbols that follow & (the equals signs, in this case).
I'm trying to learn LaTeX, currently because otherwise, my professors will be nearly unable to read my homework assignments. I've come across something I want to do, but don't seem to be able to, ie. I have searched google (possibly with a poor keyword set) and not found a solution.
The specific case is as follows: I want to put an ams flalign environment inside a box and have multiple such environments side by side. I have achieved this using minipage, but minipage asks for a width. I would like to use the smallest width in which the flalign environment fits. I realize that I can set the width to 0pt, but I can't help wondering if there's something that is intended to do this.
Also, should I be using minipage? Is there another command I don't know?
Thanks for your reply.
EDIT:
An attempted clarification as to what I want to do. I want equations which are standard, known, given, etc. and short on the left. To the right of those, I want relevant derived equations (and maybe their derivations. Further right, I want actual calculations plugged in.
I feel like what I want is a tabular environment with 3 columns, but I don't think I can put an equation environment in a tabular environment.
This looks like what I want when I render it.
\begin{minipage}[t]{0pt}
\begin{flalign*}
\sigma & = F / A&\\
A & = \pi \left(d/2\right)^2&\\
\epsilon &= \frac{\sigma}{E}&\\
\epsilon_{trans} &= - \nu \epsilon_{longi}& \\
\epsilon &= \frac{\Delta l}{l}&\\
l &= \left( \epsilon + 1 \right) \times l_0&
\end{flalign*}
\end{minipage}
\hspace*{0pt}
\begin{minipage}[t]{0pt}
\begin{flalign*}
d & = \unit[1.8]{mm} = \unit[1.8\e{-3}]{m} &\\
F_T & = \unit[1300]{N}&\\
E_{\text{stainless steel}}&=\unit[193\e9]{Pa}&\\
l_0 & = \unit[.2530]{m}&\\
\nu & = .33&\\
\sigma &= \frac{\unit[1300]{N}}{\pi \times \unit[3.24\e{-6}]{m^2}}&&= \boxed{\unit[127.7\e6]{Pa}}\\
&&&=\boxed{\unit[18,524]{psi}}\\
\epsilon &= \frac{\unit[127.7\e6]{Pa}}{\unit[193\e9]{Pa}} &&= \boxed{6.6\e{-2}}\\
\epsilon_{trans} &= -.33 \times 6.6\e{-2} &&=\boxed{-2.2\e{-2}}\\
l &= \left( 6.6\e{-2} + 1 \right) \times \unit[.2530]{m} &&= \boxed{\unit[.2797]{m}}
\end{flalign*}
\end{minipage}
I'm not sure exactly what you're trying to achieve, but amsmath's align* environment might do what you want (without resorting to minipages):
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
x&=y & X&=Y & a&=b+c & mn&=ab\\
x’&=y’ & X’&=Y’ & a’&=b & m'n'&=a'b'\\
x+x’&=y+y’ & X+X’&=Y+Y’ & a’b&=c’b & m'&=a'
\end{align*}
\end{document}
As to your minipage question: it requires a width because TeX needs to know where to break the lines. If you don't want the line-breaking algorithm to be used, you probably don't want a minipage.
Edit:
If you want multiple columns and don't care about the vertical alignment of material across the columns, that can be obtained easily enough with the multicols package:
\documentclass{article}
\usepackage{multicols}
\usepackage{lipsum}% just for some example text
\begin{document}
% The * version allows the columns to have ragged bottoms.
% The argument 2 is the number of columns.
\begin{multicols*}{2}
\lipsum[1]% one paragraph of Lorem ipsum.. filler text
\vfil% fills the remainder of the column with white space
\columnbreak% force a column break
\lipsum[2]% another paragraph of text
\vfil% fills the remainder of the column with white space
\end{multicols*}
\end{document}
You might find something to help you in the empheq and mathtools packages. empheq allows you to box equations and mathtools should provide some useful environments for stacking them horizontally.