Hence ${\Vdash}|\dot{U}|=
\lambda^+$. To show that, in ${{\bf V}}^{{{{\mathbb P}^\theta_\lambda}}}$, the algebra ${\dot{{{\mathbb B}}}^\theta_\lambda}$ is of size $\lambda^+$ it is enough to prove the following claim.

\[3.4.1\] Let $p\in{{{\mathbb P}^\theta_\lambda}}$, $j\in u^p$. Then $x_j\notin\langle x_i:i\in j\cap u^p
\rangle_{{{\mathbb B}}^p}$.

Suppose not, and let $p,j$ be a counterexample with the smallest possible ${{\rm ht}}(p)$. Necessarily, ${{\rm ht}}(p)$ is a successor ordinal, say ${{\rm ht}}(p)=\alpha+1$. So let $p=\langle\zeta^*,\tau^*, n^*,u^*,\langle
p_\xi,v_\xi:\xi<\theta\rangle\rangle$ and suppose that $v\in [u^p\cap
j]^{\textstyle <\omega}$ is such that $x_j\in \langle x_i: i\in
v\rangle_{{{\mathbb B}}^p}$. If $j\in u^*$ then $v\subseteq u^*$ and we immediately get a contradiction (applying the inductive hypothesis to $p_{\zeta^*}$). So let $\xi<\theta$ be such that $j\in u^{p_\xi}\setminus u^*$. We know that $x_j\notin\langle x_i:i\in u^*\cup (v\cap u^{p_\xi})\rangle_{{{\mathbb B}}^{p_\xi}}$ (remember clause $(\gamma)$ of the definition of $P^*_{\alpha+1}$), so we may take functions $f_0,f_1\in F^{p_\xi}$ such that $f_0{\restriction}(u^*\cup (v\cap
u^{p_\xi}))=f_1{\restriction}(u^*\cup (v\cap u^{p_\xi}))$, $f_0(j)=0$, $f_1(j)=1$. Let $g_0,g_1:u^p\longrightarrow 2$ be such that $g_\ell{\restriction}u^{p_\xi}=
f_\ell$, $g_\ell{\restriction}u^{p_\zeta}=f_0{\circ}H_{\zeta,\xi}$ for $\zeta\neq\xi$ (where $H_{\zeta,\xi}$ is the order isomorphism from $u^{p_\zeta}$ to $u^{p_\xi}$). Now one easily checks that $g_0,g_1\in F^p$ (remember the definition of the term $\sigma_{\rm maj}$). By our choices, $g_0(i)=g_1(i)$ for all $i\in v$, and $g_0(j)\neq g_1(j)$, and this is a clear contradiction with the choice 