`\documentclass[12pt,a4paper]{report}
\usepackage{amsfonts, amsmath,amsfonts,amssymb,amsthm}
\title{Zusammenfassung Analysis I + II}
\author{Ambesa}
\newcommand{\satz}[1]{\textbf{Satz #1:}\\}
\newcommand{\defi}[1]{\textbf{#1:}\\}
\newcommand{\boldd}[1]{\textbf{#1}}
\newcommand{\inr}{\in \mathbb{R}}
\newcommand{\inn}{\in \mathbb{N}}
\begin{document}
\maketitle
\newpage
\noindent
\boldd{Bem:}
\begin{itemize}
\item[i)] $X=Y \Leftrightarrow X\subset Y \wedge Y\subset X$
\item[ii)] $X \cap(Y\cupZ) = (X \wedge Y)\cup( X \wedge Z)$
\end{itemize}
\\[\baselineskip]
\satz{I.2.2.1 (de Morgan)}
Seien A, $B \subset X$ mit X \textbackslash A = $A^{C}$, X \textbackslash B = $B^{C}$. Dann gilt:
\begin{itemize}
\item[i)] $(A\cup B)^{C}=A^C \cap B^C$
\item[ii)] $(A\cap B)^{C}=A^C \cup B^C$
\end{itemize}
\\[\baselineskip]
\end{document}`
Ecco il codice!
Chieso scusa se non l’ho postato subito!
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