[Tex/LaTex] How to force alignment of “cases” brackets in an enumerate environment

amsmathcases

I am using the enumitem and amsmath packages to give the recursive definition of the valuation function $\nu$ for classical first order logic. The way I have it set up, the brackets of the cases are not aligned (due to differences in length of the expressions preceding the cases brackets).

Here is the MWE:

\documentclass[12pt]{scrartcl}
\usepackage{enumitem}
\usepackage{amsmath,amssymb}

\begin{document}

\begin{enumerate}[noitemsep,label=\alph*)]

\item $\nu(A 
\wedge B) = \begin{cases}
1,&\text{if $\nu(A) = \nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\nu(A \vee B) = \begin{cases}
1,&\text{if $\nu(A) = 1$ or $\nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\nu(A \rightarrow B) = \begin{cases}
1,&\text{if $\nu(A) = 0$ or $\nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\nu(A \leftrightarrow B) = \begin{cases}
1,&\text{if $\nu(A) = \nu(B)$};\\
0,&\text{otherwise.}\end{cases}$
\item $\nu(\forall xA) = \begin{cases}
1,&\text{if for all $d \in D, \nu(A_x(c_d)) =1$;}\\
0,&\text{otherwise.}
\end{cases}$    
\item $\nu(\exists xA) = \begin{cases}
1,&\text{if for some $d \in D, \nu(A_x(c_d)) =1$;}\\
0,&\text{otherwise.}
\end{cases}$    
\end{enumerate}

\end{document}

I would like for the brackets to be aligned down the page, ideally without creating too much ugly spacing elsewhere.

Best Answer

A trick from mathmode (In my version (2.47) it is in section 65.1 on page 113): save the longest left hand side in a box, and then output the left hand sides in a box of that length:

\documentclass[12pt]{scrartcl}
\usepackage{enumitem}
\usepackage{amsmath,amssymb}

\begin{document}
\newsavebox\lW
\sbox\lW{$\nu(A \leftrightarrow B)$}

\begin{enumerate}[noitemsep,label=\alph*)]

\item $\displaystyle\makebox[\wd\lW][r]{$\nu(A\wedge B)$} = \begin{cases}
1,&\text{if $\nu(A) = \nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\displaystyle\makebox[\wd\lW][r]{$\nu(A \vee B)$} = \begin{cases}
1,&\text{if $\nu(A) = 1$ or $\nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\displaystyle\makebox[\wd\lW][r]{$\nu(A \rightarrow B)$} = \begin{cases}
1,&\text{if $\nu(A) = 0$ or $\nu(B) = 1$};\\
0,&\text{otherwise.}\end{cases}$
\item $\displaystyle\makebox[\wd\lW][r]{$\nu(A \leftrightarrow B)$} = \begin{cases}
1,&\text{if $\nu(A) = \nu(B)$};\\
0,&\text{otherwise.}\end{cases}$
\item $\displaystyle\makebox[\wd\lW][r]{$\nu(\forall xA)$} = \begin{cases}
1,&\text{if for all $d \in D, \nu(A_x(c_d)) =1$;}\\
0,&\text{otherwise.}
\end{cases}$
\item $\displaystyle\makebox[\wd\lW][r]{$\nu(\exists xA)$} = \begin{cases}
1,&\text{if for some $d \in D, \nu(A_x(c_d)) =1$;}\\
0,&\text{otherwise.}
\end{cases}$
\end{enumerate}

\end{document}

Output

Output image

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