$q{\Vdash}{{\rm Depth}}^+(\langle\dot{a}_\xi:\xi<
\lambda^+\rangle_{{\dot{{{\mathbb B}}}^\theta_\lambda}})>\theta$, finishing the proof.

\[3.5\] Assume $\theta<\lambda$ are regular, $\lambda=\lambda^{<\lambda}$. Then ${\Vdash}_{{{{\mathbb P}^\theta_\lambda}}}{{\rm Depth}}({\dot{{{\mathbb B}}}^\theta_\lambda})=\theta$.

By Proposition \[3.4\] we know that ${\Vdash}{{\rm Depth}}^+({\dot{{{\mathbb B}}}^\theta_\lambda})>\theta$, so what we have to show is that there are no increasing sequences of length $\theta^+$ of elements of ${\dot{{{\mathbb B}}}^\theta_\lambda}$. We will show this under an additional assumption that $\theta^+ <\lambda$ (after the proof is carried out, it will be clear how one modifies it to deal with the case $\lambda=\theta^+$). Due to this additional assumption, and since the forcing notion ${{{\mathbb P}^\theta_\lambda}}$ is $(<\lambda)$–strategically closed (by Proposition \[3.3\]), it is enough to show that ${{\rm Depth}}({{\mathbb B}}^p)\leq\theta$ for each $p\in{{{\mathbb P}^\theta_\lambda}}$.

So suppose that $p\in{{{\mathbb P}^\theta_\lambda}}$ is such that ${{\rm Depth}}({{\mathbb B}}^p)\geq\theta^+$. Then we find a Boolean term $\tau$, an integer $n$ and sets $w_\rho\in [u^p]^{
\textstyle n}$ (for $\rho<\theta^+$) such that $$\rho_0<\rho_1<\theta^+\quad\Rightarrow\quad{{\mathbb B}}^p\models\tau(x_i:i\in w_{
\rho_0})<\tau(x_i:i\in w_{\rho_1}).$$ For each $\rho<\theta^+$ use Lemma \[3.5.1\] to choose a finite $p$–closed set $Z_\rho\subseteq{{\rm ht}}(p)$ containing the set $$\{\beta<{{\rm ht}}(p): (\exists j\in w_\rho)(h^p(j,\beta)<\theta)\}.$$ Look at $\Upsilon_p(Z_\rho)$ (see Definition \[UUpsilon\]). There are only $\theta$ possibilities for the values of $\Upsilon_p(Z_\rho)$, so we find $\rho_0<\rho_1<\theta^+$ such that

1.  $|Z_{\rho_0}|=|Z_{\rho_1}|$, $\Upsilon_p(Z_{\rho_0})=\Upsilon_p( 
    Z_{\rho_1})=\langle\zeta_\ell,\tau_\ell,n_\ell,\langle g_\ell,h^\ell_0,
    \ldots,h^\ell_{n_\ell-1}\rangle:\ell<k\rangle$,

2.  if $\pi^*:Z_{\rho_0}\longrightarrow Z_{\rho_1}$ 