In this code
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\title{Chiave di Basso prova}
\author{PUCK}
\date{\today}
\begin{document}
\maketitle
\section{Introduction}
Let $\raisebox{2.5mm}{\smallbassclef}\coloneqq\biggl{\raisebox{2.5mm}{\smallbassclef},\le\biggr}$ a Partial Ordered Set.
\section{Elementi di Teoria degli Insiemi}
\newtheorem{DeMorganCupCap}{Teorema di Augustus De Morgan}[subsection]
\begin{DeMorganCupCap}
Dato un insieme $X$ e due suoi sottoinsiemi, siano essi $A$ e $B$, allora valgono le seguenti due tesi:
\begin{enumerate}
\item $X\setminus(A \cup B)=(X\setminus A) \cap(X\setminus B)=\complement (A)\cap\complement(B)$;
\item $X\setminus\left(A\cap B\right)=\left(X\setminus A\right)\cup\left( X \setminus B\right)=\complement\left(A\right)\cup\complement\left(B\right)$.
\end{enumerate}
Un altro modo elegante di enunciare le Leggi di De Morgan è questo:
\begin{enumerate}
\item $\complement(A\cup B)=\complement(A)\cap\complement(B)$;
\item $\complement(A\cap B)=\complement(A)\cup\complement(B)$.
\end{enumerate}
\begin{proof}
Sia $x$ un generico elemento di $X$, allora si ha che
\begin{equation}
\begin{split}
x\in X\setminus(A\cup B)&\iff x\in X\wedge x\notin A\cup B\& \iff x\notin A\wedge x\notin B\ &\iff x\in X\setminus A=\complement (A)\wedge x\in X\setminus B= \complement(B)\ &\iff x\in \complement (A) \cap \complement(B)
\end{split}
\end{equation}
\begin{equation}
\begin{split}
x\in X\setminus(A\cap B)&\iff x\in X\wedge x\notin A\cap B\& \iff x\notin A\vee x\notin B\ &\iff x\in X\setminus A=\complement (A)\vee x\in X\setminus B= \complement(B)\ &\iff x\in \complement (A) \cup \complement(B)
\end{split}
\end{equation}
La dimostrazione è conclusa.
\end{proof}
\end{DeMorganCupCap}
\end{document}
the output is wrong
Overleaf tells me this:
How could I solv? Thank you so much
(Perhaps the order of packages is not correct, and almost of course some package is inuseful)
