The tail-flip symmetric model has no free ultrafilter on omega

First, maximality makes $U$ decide every $A\subseteq \omega$: if neither $A$ nor $\omega \setminus A$ belonged to $U$, then every $B\in U$ would meet $A$ (otherwise upward closure would put $\omega \setminus A$ in $U$), so the filter generated by $U\cup \{A\}$ would be a proper strict extension, contradicting F2. Thus exactly one of $A$ and its complement lies in $U$, since a proper filter cannot contain both.

Define $I=\{A\subseteq \omega :\omega \setminus A\in U\}$. Complementation converts upward closure to downward closure and intersections to unions, so $I$ is a proper ideal. If $A\cap B\in I$, then $(\omega \setminus A)\cup (\omega \setminus B)\in U$. Were neither complement in $U$, step 1.1 would put both $A$ and $B$ in $U$, and then $A\cap B\in U$, a contradiction. Hence $A\in I$ or $B\in I$, so $I$ is prime.

By F1 the ideal $I$ is generated by some $C\subseteq \omega$, so $I=\{A\subseteq \omega :A\subseteq C\}$. Propriety gives $C\mathrm{\ne }\omega$. Choose $k\in \omega \setminus C$. Since $I$ is prime and $\{k\}\cap (\omega \setminus \{k\})=\mathrm{\varnothing }\in I$, while $\{k\}⊈C$, we have $\omega \setminus \{k\}\in I$ and hence $\omega \setminus \{k\}\subseteq C$. Together with $k\mathrm{\notin }C$, this gives $C=\omega \setminus \{k\}$ and therefore $I=\{A\subseteq \omega :k\mathrm{\notin }A\}$. For every $B\subseteq \omega$, it follows that $B\in U$ iff $\omega \setminus B\in I$ iff $k\in B$. Thus $U$ is the principal ultrafilter at $k$ in the sense of F2. Since $U$ was arbitrary, no free ultrafilter on $\omega$ exists in the model.