I have an array that I had past help on and now I am attempting to modify that array. However, when doing so, parts of my "theorem" keep getting shifted to the next line.
Here is a MWE:
\documentclass{article}
\usepackage{amsmath,array}% http://ctan.org/pkg/{amsmath,array}
\newcommand{\twolinebrace}{\rlap{$\smash{\raisebox{.5\height}{\bigg\}}}$}}
\newlength{\LHS}\newlength{\RHS}
\newcolumntype{M}{>{$}p{\LHS}<{$}}
\newcolumntype{N}{>{$}p{\RHS}<{$}}
\begin{document}
\begin{theorem}
Let $\{A_{\alpha \colon \alpha \in A}\}$ and $\{B_{\beta \colon \beta \in B}\}$ be families of subsets if a set $X$. Then:
\[
\renewcommand{\arraystretch}{1.1}
\begin{array}{@{}l@{\quad}l@{}}
\left.\kern-\nulldelimiterspace\\
\begin{array} {>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(a) & \big(\bigcup_{\alpha\in A} A_{\alpha}\big) \cup \big(\bigcup_{\beta\in B} B_{\beta}\big) & \bigcup_{(\alpha,\beta)} A_{\alpha} \cup B_{\beta}\\
(b) & \bigg(\bigcap_{\alpha \in A A_{\alpha}} \bigg) \cap \bigg(\bigcap_{\beta \in B B_{\beta}} \bigg) & \bigcap_{(\alpha,\beta)} A_{\alpha} \cap B_{\beta}
\end{array}\right\} & {\text{Associative Laws}} \\
\left.\kern-\nulldelimiterspace\begin{array}{>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(c) & X \cup (Y \cup Z) & (X \cup Y) \cup Z \\
(d) & X \cap (Y \cap Z) & (X \cap Y) \cap Z
\end{array}\right\} & {\text{Associative Laws}} \\
\left.\kern-\nulldelimiterspace\begin{array}{>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(e) & X \cup (Y \cap Z) & (X \cup Y) \cap (X \cup Z) \\
(f) & X \cap (Y \cup Z) & (X \cap Y) \cup (X \cap Z)
\end{array}\right\} & {\text{Distributive Laws}} \\
\end{array}
\]
\end{theorem}
\end{document}
I needed to add to my array, and now I am having additional problems. Now, the "text" 'Associative Laws', 'Distributive Laws', and 'de Morgan's Laws' are not being displayed correctly. Here is my latest MWE:
\documentclass{article}
\usepackage{amsmath,array}% http://ctan.org/pkg/{amsmath,array}
\newcommand{\twolinebrace}{\rlap{$\smash{\raisebox{.5\height}{\bigg\}}}$}}
\newlength{\LHS}\newlength{\RHS}
\newcolumntype{M}{>{$}p{\LHS}<{$}}
\newcolumntype{N}{>{$}p{\RHS}<{$}}
\begin{document}
\begin{theorem}
Let $\{A_{\alpha \colon \alpha \in A}\}$ and $\{B_{\beta \colon \beta \in B}\}$ be families of subsets of a set~$X$. Then:
\[
\renewcommand{\arraystretch}{1.25}
\begin{array}{@{}l@{\quad}l@{}}
\left.\kern-\nulldelimiterspace\begin{array}{>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(a) & X \cup Y & Y \cup X \\
(b) & X \cap Y & X \cap Y
\end{array}\right\} & {\text{Commutative Laws}} \\
\left.\kern-\nulldelimiterspace\begin{array}{>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(c) & X \cup (Y \cup Z) & (X \cup Y) \cup Z \\
(d) & X \cap (Y \cap Z) & (X \cap Y) \cap Z
\end{array}\right\} & {\text{Associative Laws}} \\
\left.\kern-\nulldelimiterspace\begin{array}{>{\bfseries}p{2em}@{\quad}M@{\quad$=$\quad}N}
(e) & X \cup (Y \cap Z) & (X \cup Y) \cap (X \cup Z) \\
(f) & X \cap (Y \cup Z) & (X \cap Y) \cup (X \cap Z)
\end{array}\right\} & {\text{Distributive Laws}} \\
\end{array}
\]
\end{theorem}
Any additional help would be greatly appreciated in this matter.
\LHS
and\RHS
, haven't been assigned any values, so those lengths are effectively zero. there may be other problems as well, but zero lengths are useful only in very special circumstances, and as widths for table columns isn't one of those. – barbara beeton Nov 3 '13 at 11:25\LHS
and\RHS
are still zero width. this new example would benefit from either of the answers that have already been given. or you could\setlength{\LHS}{.75in}\setlength{\RHS}{1.5in}
and at least get something that doesn't overprint. – barbara beeton Dec 23 '13 at 20:35