2

I'm trying to make a table for some basic rules in logic, where I add each rule's name in the last column. I made all of this in the array environment:

$$\begin{array}{lcccl}
\mathrm{(a)} & \mathrm{P}                  & \mathrm{and} & \lnot(\lnot \mathrm{P})           & (\textit{Double Negation Law}) \\
\mathrm{(b)} & \mathrm{P} \lor \mathrm{Q}           & \mathrm{and} & \mathrm{Q}\lor \mathrm{P}                  & \\
\mathrm{(c)} & \mathrm{P} \land \mathrm{Q}          & \mathrm{and} & \mathrm{Q}\land \mathrm{P}                 & \\
\mathrm{(d)} & \mathrm{P}\lor (\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\lor \mathrm{Q})\lor R          & \\
\mathrm{(e)} & \mathrm{P}\land (\mathrm{Q}\land R)  & \mathrm{and} & (\mathrm{P}\land \mathrm{Q})\land R        &\\
\mathrm{(f)} & \mathrm{P}\land(\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\land \mathrm{Q})\lor(\mathrm{P}\land R) &\\
\mathrm{(g)} & \mathrm{P}\lor(\mathrm{Q}\land R)    & \mathrm{and} & (\mathrm{P}\lor \mathrm{Q})\land(\mathrm{P}\lor R)  & \\
\mathrm{(h)} & \lnot(\mathrm{P}\land \mathrm{Q})    & \mathrm{and} & \lnot \mathrm{P}\lor\lnot \mathrm{Q}       & \\
\mathrm{(i)} & \lnot(\mathrm{P}\lor \mathrm{Q})     & \mathrm{and} & \lnot \mathrm{P}\land\lnot \mathrm{Q}      & \\
\end{array}
$$

I would like to get as an end result something like this: enter image description here

3

I don't think that the array environment is the best choice here, but any ways, here is an option using the multirow package. I just add a two-row cell \multirow{2}{*}{$\left.\hbox{\rule{0cm}{.45cm}}\right\}$ De Morgan's Laws} as in the code below. The array environment is usually too dense, so, to make it more readable you can add \renewcommand{\arraystretch}{1.2} locally before the array.

\documentclass{article}
\usepackage{amsmath}
\usepackage{multirow}
\begin{document}

\begin{equation*}
\renewcommand{\arraystretch}{1.2}
\begin{array}{lcccl}
\mathrm{(a)} & \mathrm{P}                  & \mathrm{and} & \lnot(\lnot \mathrm{P})           & (\textit{Double Negation Law}) \\
\mathrm{(b)} & \mathrm{P} \lor \mathrm{Q}           & \mathrm{and} & \mathrm{Q}\lor \mathrm{P}                  & \\
\mathrm{(c)} & \mathrm{P} \land \mathrm{Q}          & \mathrm{and} & \mathrm{Q}\land \mathrm{P}                 & \\
\mathrm{(d)} & \mathrm{P}\lor (\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\lor \mathrm{Q})\lor R          & \\
\mathrm{(e)} & \mathrm{P}\land (\mathrm{Q}\land R)  & \mathrm{and} & (\mathrm{P}\land \mathrm{Q})\land R        &\\
\mathrm{(f)} & \mathrm{P}\land(\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\land \mathrm{Q})\lor(\mathrm{P}\land R) &\\
\mathrm{(g)} & \mathrm{P}\lor(\mathrm{Q}\land R)    & \mathrm{and} & (\mathrm{P}\lor \mathrm{Q})\land(\mathrm{P}\lor R)  & \\
\mathrm{(h)} & \lnot(\mathrm{P}\land \mathrm{Q})    & \mathrm{and} & \lnot \mathrm{P}\lor\lnot \mathrm{Q}       &\multirow{2}{*}{$\left.\hbox{\rule{0cm}{.45cm}}\right\}$ De Morgan's Laws} \\
\mathrm{(i)} & \lnot(\mathrm{P}\lor \mathrm{Q})     & \mathrm{and} & \lnot \mathrm{P}\land\lnot \mathrm{Q}      & \\
\end{array}
\end{equation*}

\end{document}

enter image description here

| improve this answer | |
2
\documentclass[10pt]{article}
\usepackage{amsmath}
\begin{document}

\[
\begin{array}{lcccl}
\mathrm{(a)} & \mathrm{P} & \mathrm{and} & \lnot(\lnot 
\mathrm{P})           & (\textit{Double Negation Law}) \\
\mathrm{(b)} & \mathrm{P} \lor \mathrm{Q}           & \mathrm{and} & \mathrm{Q}\lor 
\mathrm{P}                  & \\
\mathrm{(c)} & \mathrm{P} \land \mathrm{Q}          & \mathrm{and} & \mathrm{Q}\land 
\mathrm{P}                 & \\
\mathrm{(d)} & \mathrm{P}\lor (\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\lor 
\mathrm{Q})\lor R          & \\
\mathrm{(e)} & \mathrm{P}\land (\mathrm{Q}\land R)  & \mathrm{and} & (\mathrm{P}\land 
\mathrm{Q})\land R        &\\
\mathrm{(f)} & \mathrm{P}\land(\mathrm{Q}\lor R)    & \mathrm{and} & (\mathrm{P}\land 
\mathrm{Q})\lor(\mathrm{P}\land R) &\\
\mathrm{(g)} & \mathrm{P}\lor(\mathrm{Q}\land R)    & \mathrm{and} & (\mathrm{P}\lor 
\mathrm{Q})\land(\mathrm{P}\lor R)  & \\
\mathrm{(h)} & \lnot(\mathrm{P}\land \mathrm{Q})    & \mathrm{and} & \lnot 
\mathrm{P}\lor\lnot \mathrm{Q}       &  
\makebox(0,0){\put(0,-20){%
  \left.\rule{0pt}{1.06\normalbaselineskip}\right\}\text{De Morgan's laws}}}\\  
\mathrm{(i)} & \lnot(\mathrm{P}\lor \mathrm{Q})     & \mathrm{and} & \lnot 
\mathrm{P}\land\lnot \mathrm{Q}      & 
\end{array}
\]

\end{document}

enter image description here

| improve this answer | |
  • Thanks for your answer! I accepted Abo Ammar's one since it solves the little spacing problem. – Workaholic Feb 24 '16 at 21:43
1

Here is one option that provide an actual list (rather than unbreakable block/array):

enter image description here

\documentclass{article}

\usepackage{enumitem}

\newlength{\leftboxlen}
\newcommand{\setleftbox}[1]{\settowidth{\leftboxlen}{#1}}
\newcommand{\leftbox}[2][c]{\makebox[\leftboxlen][#1]{#2}}
\newlength{\rightboxlen}
\newcommand{\setrightbox}[1]{\settowidth{\rightboxlen}{#1}}
\newcommand{\rightbox}[2][c]{\makebox[\rightboxlen][#1]{#2}}

\begin{document}

\noindent\textbf{Theorem 1.6.}
\setleftbox{$\mathrm{P} \land (\mathrm{Q} \land R)$}%
\setrightbox{$(\mathrm{P} \land \mathrm{Q}) \lor (\mathrm{P} \land R)$}%
\begin{enumerate}[label=(\alph*),nosep]
  \item \leftbox{$\mathrm{P}$} and \rightbox{$\lnot(\lnot \mathrm{P})$} \qquad (\textit{Double Negation Law})
  \item \leftbox{$\mathrm{P} \lor \mathrm{Q}$} and \rightbox{$\mathrm{Q} \lor \mathrm{P}$}
  \item \leftbox{$\mathrm{P} \land \mathrm{Q}$} and \rightbox{$\mathrm{Q} \land \mathrm{P}$}
  \item \leftbox{$\mathrm{P} \lor (\mathrm{Q} \lor R)$} and \rightbox{$(\mathrm{P} \lor \mathrm{Q}) \lor R$}
  \item \leftbox{$\mathrm{P} \land (\mathrm{Q} \land R)$} and \rightbox{$(\mathrm{P} \land \mathrm{Q}) \land R$}
  \item \leftbox{$\mathrm{P} \land(\mathrm{Q} \lor R)$} and \rightbox{$(\mathrm{P} \land \mathrm{Q}) \lor (\mathrm{P} \land R)$}
  \item \leftbox{$\mathrm{P} \lor (\mathrm{Q} \land R)$} and \rightbox{$(\mathrm{P} \lor \mathrm{Q}) \land (\mathrm{P} \lor R)$}
  \item \leftbox{$\lnot (\mathrm{P} \land \mathrm{Q})$} and \rightbox{$\lnot \mathrm{P} \lor \lnot \mathrm{Q}$} \qquad
    \raisebox{-.45\height}[0pt][0pt]{$\left.\kern-\nulldelimiterspace\begin{array}{@{}c@{}} \mathstrut \\ \mathstrut \end{array}\right\} \mbox{(\textit{De Morgan's Law})}$}
  \item \leftbox{$\lnot(\mathrm{P} \lor \mathrm{Q})$} and \rightbox{$\lnot \mathrm{P} \land \lnot \mathrm{Q}$}
\end{enumerate}

\end{document}

Horizontal alignment of the structure is achieved using boxes. The left section is set within \leftbox (which has a width set through \setleftbox), while the right section is set within \rightbox (and a similarly names \setrightbox).

The De Morgan's Law notation is a lowered stack (2-row array) with zero height/depth.

| improve this answer | |

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