\resizebox within \begin{align}... \end{align} not working while using \intertext{

Question

I have some very wide equations within a proof environment which I want to contain in a two-column document, I thus think of using \resizebox{1.3\columnwidth}{!}{% to archive that generate this error message "package amsmath error: invalid use of \intertext{" as seen in MWE1 below.

\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc}

\usepackage{multicol}% set this to 2 colomn
\setlength{\columnsep}{7mm}% set the space betw colomn 7mm

\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[left=2.00cm, right=2.00cm, top=2.00cm, bottom=2.00cm]{geometry}

\usepackage{ulem}
\newcommand{\msout}[1]{\text{\sout{\ensuremath{#1}}}}

\newcommand{\numberthis}{\refstepcounter{equation}\tag{\arabic{equation}}}
\newcommand{\labeln}[1]{\numberthis\label{#1}}
\newcommand{\mathonehalf}{\ensuremath{\frac{1}{2}}}

\usepackage{blindtext}
\begin{document}
\begin{multicols}{2}
\blindtext
\blindtext
\begin{proof}
\resizebox{1.3\columnwidth}{!}{%
$\begin{aligned}
\vartriangle Y=& \dfrac{n!}{\left(n-X-1\right)!\,X+1!}\left(\tfrac{1}{2}
n\right)\left(\tfrac{1}{2} \sqrt{n}\right) \nonumber\\
&\,- \dfrac{n!}{\left(n-X\right)!\,X!}\left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right) \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)n!\nonumber\\
&\,\left[\dfrac{\left(n-x\right)!\,x!-\left(n-x-1\right)!\left(n-x\right)!}{\left(n-x-1\right)!\left(n-x\right)!\,\left(n-x\right)!\,x!}\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2}
\sqrt{n}\right)n!\nonumber\\
&\,\left[\dfrac{\msout{\mathsf{\left(n-x-1\right)!\,x!}}\left[\left(n-x\right)-\left(x+1\right)\right]}{\msout{\mathsf{\left(n-x-1\right)!}}\left(x+1\right)!\,\left(n-x\right)!\,\msout{\mathsf{x!}}}\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)\left[\dfrac{n!}{\left(x+1\right)!\,\left(n-x\right)!}\right]
\nonumber\\
&\,\left[\left(n-x\right)-\left(x+1\right)\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)
\left[\dfrac{n!}{\left(n-x\right)!\,x!}\right]\nonumber\\
&\,\dfrac{\left(n-x-x-1\right)}{\left(x+1\right)} \nonumber\\
\intertext{From equation \ref{eq:dz} and equation \eqref{eq:sigma} }\\
\end{aligned}$ 
}
\end{proof}

References to $\sigma$ \eqref{eq:sigma} and $\Delta Z$ \eqref{eq:dz}.
\blindtext
\end{multicols}
\end{document}

Even when I use \begin{align}...\end{align} it shows that the maths is not in maths mode as seen in MWE2 below.

\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc}

\usepackage{multicol}% set this to 2 colomn
\setlength{\columnsep}{7mm}% set the space betw colomn 7mm

\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[left=2.00cm, right=2.00cm, top=2.00cm, bottom=2.00cm]{geometry}

\usepackage{ulem}
\newcommand{\msout}[1]{\text{\sout{\ensuremath{#1}}}}

\newcommand{\numberthis}{\refstepcounter{equation}\tag{\arabic{equation}}}
\newcommand{\labeln}[1]{\numberthis\label{#1}}
\newcommand{\mathonehalf}{\ensuremath{\frac{1}{2}}}

\usepackage{blindtext}
\begin{document}
\begin{multicols}{2}
\blindtext
\blindtext
\begin{proof}
\resizebox{1.3\columnwidth}{!}{%
\begin{align}
\vartriangle Y=& \dfrac{n!}{\left(n-X-1\right)!\,X+1!}\left(\tfrac{1}{2}
n\right)\left(\tfrac{1}{2} \sqrt{n}\right) \nonumber\\
&\,- \dfrac{n!}{\left(n-X\right)!\,X!}\left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right) \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)n!\nonumber\\
&\,\left[\dfrac{\left(n-x\right)!\,x!-\left(n-x-1\right)!\left(n-x\right)!}{\left(n-x-1\right)!\left(n-x\right)!\,\left(n-x\right)!\,x!}\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2}
\sqrt{n}\right)n!\nonumber\\
&\,\left[\dfrac{\msout{\mathsf{\left(n-x-1\right)!\,x!}}\left[\left(n-x\right)-\left(x+1\right)\right]}{\msout{\mathsf{\left(n-x-1\right)!}}\left(x+1\right)!\,\left(n-x\right)!\,\msout{\mathsf{x!}}}\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)\left[\dfrac{n!}{\left(x+1\right)!\,\left(n-x\right)!}\right]
\nonumber\\
&\,\left[\left(n-x\right)-\left(x+1\right)\right] \nonumber\\
=& \left(\tfrac{1}{2} n\right)\left(\tfrac{1}{2} \sqrt{n}\right)
\left[\dfrac{n!}{\left(n-x\right)!\,x!}\right]\nonumber\\
&\,\dfrac{\left(n-x-x-1\right)}{\left(x+1\right)} \nonumber\\
\intertext{From equation \ref{eq:dz} and equation \eqref{eq:sigma} }\\
\end{align}
}
\end{proof}

References to $\sigma$ \eqref{eq:sigma} and $\Delta Z$ \eqref{eq:dz}.
\blindtext
\end{multicols}
\end{document}

I want to use texts within the proof environment and also want to use the\resizebox{ command while I keep the mathematical part align but not aliging the text part.

Answer 1

No need to use the \resizebox sledge hammer. Just get rid of all 39 pairs of \left and \right, and you'll save a surprising amount of space. Use \bigl, \bigr, \biggl and \biggr only where needed. By using an align* environment, you can get rid of all those \nonumber directly.

Incidentally, your code sometimes uses x, and sometimes it uses X. Is there a difference?

\documentclass[12pt,a4paper]{article}
\usepackage{multicol}% set this to 2 colomn
\setlength{\columnsep}{7mm}% set the space between columns to 7mm

\usepackage{amsmath,amsthm,amssymb}
%\allowdisplaybreaks  % use only if necessary
\usepackage[margin=2cm]{geometry}

\usepackage{ulem}
\newcommand{\msout}[1]{\text{\sout{\ensuremath{#1}}}}

\newcommand{\numberthis}{\refstepcounter{equation}\tag{\arabic{equation}}}
\newcommand{\labeln}[1]{\numberthis\label{#1}}
\newcommand{\mathonehalf}{\ensuremath{\frac{1}{2}}}

\usepackage{blindtext}
\begin{document}

\begin{multicols}{2}
\blindtext
\blindtext
\begin{proof}
\begin{align*}
\Delta Y
&= \frac{n!}{(n-X-1)!\,X+1!}
   \bigl(\tfrac{1}{2}n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr) \\
&\quad - \frac{n!}{(n-X)!\,X!}
   \bigl(\tfrac{1}{2} n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr) \\
&= \bigl(\tfrac{1}{2} n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr) n!\\
&\quad\times\biggl[\frac{(n-x)!\,x!-(n-x-1)!\,(n-x)!}{(n-x-1)!\,(n-x)!\,(n-x)!\,x!}\biggr] \\
&= \bigl(\tfrac{1}{2} n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr) n!\\
&\quad\times\biggl[\frac{\msout{\mathsf{(n-x-1)!\,x!}}[(n-x)-(x+1)]}{\msout{\mathsf{(n-x-1)!}}\,(x+1)!\,(n-x)!\,\msout{\mathsf{x!}}}\biggr] \\
&= \bigl(\tfrac{1}{2} n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr) \sqrt{n})
   \biggl[\frac{n!}{(x+1)!\,(n-x)!}\biggr] \\
&\quad\times[(n-x)-(x+1)] \\
&= \bigl(\tfrac{1}{2} n\bigr)\bigl(\tfrac{1}{2} \sqrt{n}\,\bigr)
   \biggl[\frac{n!}{(n-x)!\,x!}\biggr]\\
&\quad\times\frac{(n-x-x-1)}{(x+1)} 
\end{align*}
from equations \eqref{eq:dz} and \eqref{eq:sigma}.
\end{proof}

References to $\sigma$ \eqref{eq:sigma} and $\Delta Z$ \eqref{eq:dz}.

\end{multicols}
\end{document}