\documentclass{article}

\usepackage{1231num}
\usepackage{amsthm}
\usepackage{amssymb}

\newtheorem*{eg}{Example}

\newcommand{\ZZ}{\mathbb Z}
\newcommand{\fain}[2]{\forall{#1\in#2}\ \ }
\let\leq\leqslant
\let\geq\geqslant


\begin{document}
\begin{eg}
$1+2+\dots+n=\frac12\,n(n+1)$ for all $n\in\ZZ_{\geq1}$.
\end{eg}

\begin{proof}
\begin{numpf}
\pfln For each $n\in\ZZ_{\geq1}$,
 let $P(n)$ be the proposition ``\,$1+2+\dots+n=\frac12\,n(n+1)$\,''.
\pfln (Base step)\enspace
 $P(1)$ is true because $1=\frac12\times1\times(1+1)$.
\pfln (Induction step)
 \begin{subpf}
 \pfln Let $k\in\ZZ_{\geq1}$ such that $P(k)$ is true,
  i.e.,~such that
  \begin{equation*}
   1+2+\dots+k=\frac12\,k(k+1).
  \end{equation*}\pfalignedjot\jot
 \begin{pfaligned}{Then}
   &1+2+\dots+k+(k+1)\\
   &=\frac12\,k(k+1)+(k+1)
   &&\text{by the induction hypothesis $P(k)$;}\\
   &=\Bigl(\frac k2+1\Bigr)(k+1)
    =\frac{k+2}2\,(k+1)\\
   &=\frac12\,(k+1)((k+1)+1).
 \end{pfaligned}
 \pfln So $P(k+1)$ is true.
 \end{subpf}
\pfln Hence $\fain n{\ZZ_{\geq1}}{P(n)}$ is true by MI. \qedhere
\end{numpf}
\end{proof}
\end{document}