How to put therefore and implies symbols

Question

\documentclass{article}
\usepackage{graphicx}

\begin{document}
\vspace{\baselineskip}\noindent
\textbf{THEOREM :} If an operator has both Left Identity and Right Identity then it is \emph{UNIQUE}.
\vspace{\baselineskip}\noindent
\textbf{PROOF :} Let e_{l} is left identity 
                         therefore e_{l} * e_{r} 
                             this implies e_{r}
\end{document}

Answer 1

As stated in the comments, you get the symbols in mathmode simply by writing them down. Packages like amsmath and amssymb support you.

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{theorem}{THEOREM}
\newtheorem{proof}{PROOF}
\begin{document}
\begin{theorem}
If an operator has both Left Identity and
Right Identity then it is \emph{UNIQUE}.
\end{theorem}
\begin{proof}
Let $e_{l}$ is left identity 
$\therefore e_{l} * e_{r} \implies e_{r}$
\end{proof}
\end{document}

Answer 2

A somewhat larger version of \therefore may be built as:

\dot{.\hspace{.095in}.}\hspace{.5in}

Comparison traditional and larger version:

Answer 3

Maybe you could simply use lualatex and a font which actually has the character:

%!TEX TS-program = lualatex
\documentclass{article}
\usepackage{fontspec}
    \setmainfont{Brill}% just an example -- download at http://www.brill.com/about/brill-fonts
\newtheorem{theorem}{THEOREM}
\newtheorem{proof}{PROOF}
\begin{document}
\begin{theorem}
If an operator has both Left Identity and
Right Identity then it is \emph{UNIQUE}.
\end{theorem}
\begin{proof}
Let $e_{l}$ is left identity 
∴ $e_{l} * e_{r}$ ⇒ $e_{r}$
\end{proof}
\end{document}

Answer 4

Initialisation Code:

\def\therefore{\boldsymbol{\text{ }
\leavevmode
\lower0.4ex\hbox{$\cdot$}
\kern-.5em\raise0.7ex\hbox{$\cdot$}
\kern-0.55em\lower0.4ex\hbox{$\cdot$}
\thinspace\text{ }}}

And can then be called on using:

\therefore

Which renders as:

Answer 5

Not entirely happy with any of the handmade therefores proposed above, I thought would offer this one also, which in my opinion has a good balance of dot size (in between \bullet and \cdot) and spacing:

\def\therefore{{\tiny$\bullet$}\kern-0.2ex\raisebox{1ex}{\tiny$\bullet$}\kern-0.2ex{\tiny$\bullet$}}