Parenthesis within arrays

Question

I use Typora. I have obtained the following, using an array and multiple quads :

$\begin{array}{lcl} P_4 & = & P_1 & \cup & \\ 
& \cup & P_2 & \cup & \\ 
& \cup & \left \{ S \xrightarrow[(1)]{} b \quad 
    \mid \quad S_1 \rightarrow b \in P_1, 
         \quad b \in X \cup \left \{ \lambda \right \} 
         \quad \right \} &\cup & \\ 
&\cup & \left \{ S \xrightarrow[(2)]{} bC \quad 
   \mid \quad S_1 \rightarrow bC \in P_1, 
        \quad b \in X, \quad b \ne \lambda, 
        \quad C \in V_1 \quad \right \} 
\end{array}$

instead, I would want to obtain the following:

I have tried, but, I obtain an error if I use parenthesis within array. Please, can you tell me, how can I obtain that tabular positioning? Thanks! :)

Answer 1

Here's a deliberately differently-styled answer, which uses a single align* environment and no array environments. Observe that the \cup symbols at the starts of rows 2, 3, and 4 are shifted to the right of the = symbol in row 1, to emphasize that p (at the start of row 1) is the union of the remaining stuff on all four lines. With such a layout, it's also not necessary to provide \cup symbols at the ends of rows 1, 2, and 3.

Observe also that I use \Bigm| instead of \mid in rows 3 and 4, to provide some more visual support for what the main things in those two rows are. I also use simple spaces instead of \quad, as I believe the latter inserts too much horizontal displacement.

\documentclass{article} 
\usepackage{amsmath} % for 'align*' env.
\begin{document}
\begin{align*}
P_4 &= P_1 \\
    &\quad \cup P_2 \\
    &\quad \cup\Bigl\{ S\xrightarrow[(1)]{} b\phantom{C} 
               \Bigm|  S_1\to b\phantom{C}\in P_1,\  
                       b\in X\cup\{\lambda\} 
               \Bigr\} \\
    &\quad \cup\Bigl\{ S\xrightarrow[(2)]{}bC 
               \Bigm|  S_1\to bC\in P_1,\ 
                       b\in X,\ 
                       b\notin\lambda,\ C\in V_1
               \Bigr\}
\end{align*}
\end{document}

Answer 2

I would exploit having a short second term:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{equation*}
\begin{aligned}
P_4 = P_1
& \cup P_2 \\
& \cup \Bigl\{ S \xrightarrow[(1)]{} b\hphantom{C} \Bigm|
         S_1 \rightarrow b \in P_1, b \in X \cup \{\lambda\}
       \Bigr\} \\
& \cup \Bigl\{ S \xrightarrow[(2)]{} bC \Bigm|
         S_1 \rightarrow bC \in P_1,
          b \in X,  b \ne \lambda,
          C \in V_1
       \Bigr\}
\end{aligned}
\end{equation*}

\end{document}

The trailing \cup symbols do nothing and appear as hanging from nowhere. I see no real reason for artificially aligning unconnected bits, except for the vertical bar. To the contrary, the asymmetry between the terms will help the reader in spotting the differences.

Answer 3

One way to accomplish this is using a \hphantom.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
$\begin{array}{lcll} P_4 & = & P_1 & \cup  \\ 
& \cup & P_2 & \cup  \\ 
& \cup & \left \{ 
\makebox[0pt][l]{$S \xrightarrow[(1)]{} b \quad$}
\hphantom{S \xrightarrow[(2)]{} bC \quad}
 \mid \makebox[0pt][l]{$\quad S_1 \rightarrow b \in
P_1, \quad b \in X \cup \left \{ \lambda \right \} \quad$}
\hphantom{\quad S_1 \rightarrow bC \in P_1, \quad b \in X, \quad b \ne \lambda, \quad C \in V_1 \quad} \right \} &\cup  \\
&\cup& \left \{ S \xrightarrow[(2)]{} bC \quad \mid \quad S_1 \rightarrow bC \in P_1, \quad b \in X, \quad b \ne \lambda, \quad C \in V_1 \quad \right \} 
\end{array}$
\end{document}

Alternative: \vphantoms

\documentclass{article}
\usepackage{amsmath}
\begin{document}
$\begin{array}{lclclll} P_4 & = & P_1 & & & & \cup  \\ 
& \cup & P_2 & & & & \cup  \\ 
& \cup & \left \{ 
S \xrightarrow[(1)]{} b \right.&  \mid &  S_1 \rightarrow b \in
P_1, \quad b \in X \cup \left \{ \lambda \right \}& 
\left.\vphantom{S \xrightarrow[(1)]{} b}\right\} &\cup  \\
&\cup& \left\{ S \xrightarrow[(2)]{} bC \right.& \mid 
& S_1 \rightarrow bC \in P_1, \quad b \in X, \quad b \ne \lambda, \quad C \in V_1 
& \left.\vphantom{S \xrightarrow[(1)]{} b}\right\}
\end{array}$
\end{document}

But do you really like the outcome? (To me it looks almost as terrible as a pineapple on a pizza ;-)

Answer 4

It's as simple to do that with alignat, and you have full control on the spacing between the element of the formula:

 \documentclass{article}

\usepackage{mathtools}
\usepackage{makebox} 

\begin{document}

\begin{alignat*}{6}
  P_4 & \makebox*{${}\cup{}$}{$ = $} P_1\\
  &\cup P_2 & & & & & & & & & & \cup\\
 & \cup \Bigl\{S \xrightarrow[(1)]{} b & & \mid S \xrightarrow{} b\in P_1, & \enspace & b\in X\cup\{\lambda\} & \enspace & & & \Bigr\} &\enspace & \cup\\
 & \cup \Bigl\{S \xrightarrow[(2)]{} b C& & \mid S \xrightarrow{} b\in P_1, &\enspace & b\in X ,& \enspace & b\ne\lambda, & \enspace C\in V_1&\Bigr\} & &\cup
\end{alignat*}

\end{document} 

Answer 5

Adapting this answer by Mico, we can produce the following:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\setlength{\arraycolsep}{3pt}
\begin{array}{rcl c l c l c r c r c }
P_4 & = &    P_1  &&&&&&&&& \cup \\
    & \cup & P_2  &&&&&&&&& \cup \\
    & \cup & \Bigl\{ S \xrightarrow[(1)]{} b & \mid & S_1 \to b  \in P_1, && b \in X \cup  \{ \lambda \} &&                  && \Bigr\}            & \cup \\
    & \cup & \Bigl\{ S \xrightarrow[(2)]{} bC & \mid & S_1 \to bC \in P_1, && b \in X,                    &&   b\neq \lambda, && C\in V_1   \Bigr\} & \cup 
\end{array}
\]
\end{document}