principals tell everything they know to each other. Therefore, $(P-\{o\})\times Knowledge(k_0)$ should be included in the maximal state of knowledge. This is equivalent to assuming that each honest principal draws $Knowledge(k_0)$ because $k\subseteq f^*_R(k_0)$ implies that $f^*_R(k_0)=f^*_R(k_0\cup k)$.

\[arrow\] Let $[(P-\{o\})\times V_0]\subseteq k_0$ with $V_0=Knowledge(k_0)$ and let $k_n=f^n_R(k_0)$. Then there exists a unique set $X_n\subseteq V$ such that $$k_n=[(P-\{o\})\times V_0] \cup [\{o\}\times X_n].\label{xn}$$ The set $X_n$ has the property that $X_n=g^n_{R,k_0}(X_0)$.

[*Proof.*]{}

We use induction on $n$. For $n=0$, $X_n=X_0=g^n_{R,k_0}(X_0)$. Since $(P-\{o\})\times V_0\subseteq k_0$ and $V_0=Knowledge(k_0)$, there exists a unique $X_0$ such that $k_0$ satisfies (\[xn\]).

Let $X_n=g^n_R(X_0)$ be a unique solution to (\[xn\]) and\
$Knowledge(k_n)=V_0$ (our induction hypothesis). We know that $k_{n+1}=f_R(k_n)$ and, therefore, $Knowledge(k_{n+1})=Knowledge(k_n)=V_0$. Together with $[(P-\{o\})\times V_0] \subseteq k_n\subseteq k_{n+1}$, this implies the existence a unique $X_{n+1}$ for which $k_{n+1}$ satisfies (\[xn\]). We now need to prove that $X_{n+1}=g^{n+1}_{R,k_0}(X_0)$.

Definition (\[xn\]) lets us infer that $x\in X_{n+1} \Longleftrightarrow (o,x)\in k_{n+1}=f_R(k_n)$. According to Definition (\[fR-def\]), $(o,x)\in f_R(k_n)$ if and only if i) $(o,x)\in k_n$, which is, by (\[xn\]), equivalent to $x\in X_n$, or ii) there exists a $p\in P$ and $k_\sigma\in K$ such that $(p,x)\in k_n$, $k_\sigma\subseteq k_n$, and $(p,x,o,k_\sigma)\in R$. Since there are no self-rules $(o,v,o,k_\sigma)\in R$, 