Difficulty modifying an array environment

Question

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.

Answer 1

Perhaps the following is a little cleaner:

\documentclass{article}
\usepackage{amsmath,amsthm,array}% http://ctan.org/pkg/{amsmath,amsthm,array}
\newcommand{\twolinebrace}{\left.\kern-\nulldelimiterspace\begin{array}{@{}c@{}}\\\\\end{array}\right\}}
\newtheorem{theorem}{Theorem}
\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:
\[
  \renewcommand{\arraystretch}{1.25}
  \begin{array}{@{}l@{\quad}l@{}}
    \begin{array}{>{\bfseries}p{2em}@{\quad}r@{{}={}}l}
      (a) & 
        \big(\bigcup_{\alpha\in A} A_{\alpha}\big) \cup \big(\bigcup_{\beta\in B} B_{\beta}\big) & % LHS
        \bigcup_{(\alpha,\beta)} A_{\alpha} \cup B_{\beta} \\ % RHS
      (b) & 
        \big(\bigcap_{\alpha \in A} A_{\alpha} \big) \cap \big(\bigcap_{\beta \in B} B_{\beta} \big) & % LHS
        \bigcap_{(\alpha,\beta)} A_{\alpha} \cap B_{\beta} % RHS
    \end{array} & \twolinebrace\text{Associative Laws} \\
    \begin{array}{>{\bfseries}p{2em}@{\quad}r@{{}={}}l}
      (c) & 
        X \cup (Y \cup Z) & % LHS
        (X \cup Y) \cup Z \\ % RHS
      (d) & 
        X \cap (Y \cap Z) & % LHS
        (X \cap Y) \cap Z % RHS
    \end{array} & \twolinebrace\text{Associative Laws} \\
    \begin{array}{>{\bfseries}p{2em}@{\quad}r@{{}={}}l}
      (e) & 
        X \cup (Y \cap Z) & % LHS
        (X \cup Y) \cap (X \cup Z) \\ % RHS
      (f) & 
        X \cap (Y \cup Z) & % LHS
        (X \cap Y) \cup (X \cap Z) % RHS
    \end{array} & \twolinebrace\text{Distributive Laws}
  \end{array}
\]
\end{theorem}
\end{document}

Note that using an array like this (or even just setting the content in a display \[...\]) it can't be broken across the page boundary. If you wish to overcome this limitation, you'll have to set the display as a list rather, or break it up into separate chunks containing each "law pair", allowing content to break inbetween.

Answer 2

Just a different flavor to the above though I do not recommend it for multiple uses. The reason being that you would have to adjust the labels each time. But anyhow here it is:

\documentclass{article}
\usepackage{amsmath,amsthm}% http://ctan.org/pkg/{amsmath,array}
\usepackage{picture}
\usepackage{enumitem}
\newtheorem{theorem}{Theorem}
\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:
\begin{enumerate}[label=\bfseries(\alph*)]
    \item $\big(\bigcup_{\alpha\in A} A_{\alpha}\big) \cup \big(\bigcup_{\beta\in B} B_{\beta}\big) = \bigcup_{(\alpha,\beta)} A_{\alpha} \cup B_{\beta}$
    \item $\big(\bigcap_{\alpha \in A} A_{\alpha} \big) \cap \big(\bigcap_{\beta \in B} B_{\beta} \big)=\bigcap_{(\alpha,\beta)} A_{\alpha} \cap B_{\beta}$
                \makebox(0,0){\put(0,10ex){%
               $\left.\rule{0pt}{5ex}\right\}$ Associative Laws}}
    \item $X \cup (Y \cup Z)= (X \cup Y) \cup Z$  
  \item $X \cap (Y \cap Z) = (X \cap Y) \cap Z$
                \makebox(0,0){\put(7em,10ex){%
               $\left.\rule{0pt}{5ex}\right\}$ Associative Laws}}
  \item $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$
  \item $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$
                \makebox(0,0){\put(4.25em,10ex){%
               $\left.\rule{0pt}{5ex}\right\}$ Distributive Laws}}
\end{enumerate}
\end{theorem}
\end{document}