# Global alignment

I'm trying to do some sort of global alignment for my equations and didn't succeed to find anything online. More precisely, my cases environment inside my aligned and equation* has its own & alignment-symbols that I want to be working with other ones inside aligned. Hopefully this MWE (using amsmath) makes it understandable:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}

\begin{document}

\begin{equation*}
\begin{aligned}
& \left| f_{(x)} \right| &\overset{!}{\leq} &L \left| x - y \right| &                      \\
\Leftrightarrow & \left| x - y \right|   &\leq              &L \left| x - y \right| &                      \\
\Leftrightarrow & \begin{cases}
x - y                  &\leq              &L (x - y)              &\text{ for } x \geq y \\
y - x                  &\leq              &L (y - x)              &\text{ for } x > y    \\
\end{cases}                                                                              \\
\end{aligned}
\end{equation*}

\end{document}


How do I align every part over each other like the &s in the code above?

Thank you for any hints. I'm also happy for feedback on my code-style :)

Extra question: Why do I get an "Extra alignment tab has been changed to \cr." and an "underfull hbox" at the very end?

• It's not quite a MWE. A MWE is to be typeset "out of the box", so it must have \documentclass{}, the necessary packages, \begin{document} and so on… – Franck Pastor Dec 16 '17 at 14:00
• @FranckPastor done, thanks for pointing out! – 4ster Dec 16 '17 at 14:54
• The cases environment defines a box of its own IIRC, so I can't see how it can be managed this way. By the way, this environment doesn't accept more than one ampersand per line, so your code can't be typeset as it is. – Franck Pastor Dec 16 '17 at 15:20
• Methinks you are trying too align too much. Is that really necessary? – Harald Hanche-Olsen Dec 16 '17 at 16:02
• @FranckPastor, again, thank you. That will probably mean that I'm dropping the use of cases then and have to go for doing it by myself... – 4ster Dec 16 '17 at 17:53

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}

\begin{document}

\begin{equation*}
\begin{alignedat}{3}
&& | f_{(x)}| &\overset{!}{\leq} L | x - y | &                      \\
&{\Leftrightarrow} & | x - y | &\leq          L | x - y | &                      \\
&\smash{\raisebox{-8pt}{$\Leftrightarrow\biggl\{$}}
&  x - y    &\leq   L (x - y)   &\text{ for } x \geq y \\
&&y - x     &\leq   L (y - x)   &\text{ for } x > y
\end{alignedat}
\end{equation*}

\end{document}

• Wow this just looks perfect as imagined. After reading about alignedat and learning about cases only supporting one ampersand, I guess doing it all manually is the only option. However, and don't get me wrong on this, I really don't like raisebox and biggl\{ instead of dynamical parantheses like \left\{. I guess nesting two multline boxes will be the way to go... – 4ster Dec 16 '17 at 18:00
• @4ster it's almost always better to choose the size than use left/right eg uses such as \left| x - y \right| should be avoided. – David Carlisle Dec 16 '17 at 18:03
• that's interesting to hear. I thought keeping everything variable (e.g. allowing to change the documents font size) was a good measure to take - 8pt in your code didn't seem to comply with that approach. Now I guess I need a good guide for LaTeX-style to change my approach. With your experience, can you recommend something where I can get a deeper understanding of this? – 4ster Dec 16 '17 at 18:48
• @4ster see tex.stackexchange.com/q/173717/1090 – David Carlisle Dec 16 '17 at 19:23

With some gentle persuasion applied to bigdelim:

\documentclass{article}
\usepackage{amsmath}
\usepackage{array,bigdelim}
\usepackage{etoolbox}

\newcommand{\Nleq}{\overset{!}{\leq}}

\makeatletter
\newcommand{\reducedelim}{%
\patchcmd{\@ldelim}{\multirow@dima}{0.75\multirow@dima}{}{}%
}
\makeatother

\begin{document}

\begin{equation*}
\renewcommand{\arraystretch}{1.5}
\setlength{\arraycolsep}{0pt}
\begin{array}{ l r >{{}}c<{{}} l @{\qquad} l }
& \lvert f_{(x)} \rvert &\Nleq& L \lvert x - y \rvert \\
\Leftrightarrow
& \lvert x - y \rvert  &\leq&  L \lvert x - y \rvert \\
\reducedelim\ldelim\lbrace{2}{*}[$\Leftrightarrow$\qquad]
& x - y                 &\leq&  L (x - y) &\text{for $x \geq y$} \\
& y - x                 &\leq&  L (y - x) &\text{for $x > y$}
\end{array}
\end{equation*}

\end{document}


On the other hand, the middle inequality is satisfied, for all x and y, if and only if L ≥ 1.