\documentclass{article}

\begin{document}

\begin{center}
  {\bf The theorem of Cauchy and Davenport}
\end{center}

{\bf Theorem.} If $A$, abd $B$ are in $Z/p\bf Z$, then
$card(A+B)\geq min(p,A+B-1$).

Proof. Indzuction on the size of $B$, and we can suppose that $0$ is in $B$. 

For any $b$, it is not possible that $A+b=A$, except for $b=0$. So we
have some $a+b\notin A$. The set $X=\{b';a+b'\in A\}$ is smaller
than~$B$, and the set $Y=A$ union with $a+(B-X)$ is larger than $A$.

Induction gives $card(X+Y)$ at least $min(p,X+Y-1)$, which is what we want. 

Here's an example show we cannot do better: take $A=B=0,1,...,n-1$.

\end{document}
