Let $i^x_\xi=\min(u^{q^x_\xi}\setminus u^x)$. Then, as $q^x_\xi$ and $q^x_\zeta$ are isomorphic and the isomorphism is the identity on $u^x$, we have $(\forall\ell<k)(h^p(i^x_\xi,\alpha_\ell^x)=g_k(\ell))$. Hence $\pi_\xi
(i^0_\xi)=i^1_\xi$, and therefore $\pi_\xi[u^0\cap U[q^0_\xi,Z_0\setminus
\{\alpha^0_k\}]]=u^1\cap U[q^1_\xi,Z_1\setminus\{\alpha^1_k\}]$. But since the mappings $\pi_\xi$ are order preserving, the last equality implies that $\pi_\xi\restriction (u^0\cap U[q^0_\xi,Z_0\setminus\{\alpha^0_k\}])=
\pi_\zeta\restriction (u^0\cap U[q^0_\zeta,Z_0\setminus\{\alpha^0_k\}])$, and hence $\pi=\bigcup\limits_{\xi<\theta}\pi_\xi$ is a function, and it is an order isomorphism from $U[q_0,Z_0]=U[p,Z_0]$ onto $U[q_1,Z_1]=U[p,Z_1]$ satisfying $(\otimes)$.

The algebra and why it is OK (in ${{\bf V}}^{{{{\mathbb P}^\theta_\lambda}}}$)
==============================================================================

Let ${\dot{{{\mathbb B}}}^\theta_\lambda}$ and $\dot{U}$ be ${{{\mathbb P}^\theta_\lambda}}$–names such that $${\Vdash}_{{{{\mathbb P}^\theta_\lambda}}}\mbox{`` }{\dot{{{\mathbb B}}}^\theta_\lambda}=\bigcup\{{{\mathbb B}}^p:p\in \Gamma_{{{{\mathbb P}^\theta_\lambda}}}\}\mbox{
''}\quad\mbox{ and }\quad{\Vdash}_{{{{\mathbb P}^\theta_\lambda}}}\mbox{`` }\dot{U}=\bigcup\{u^p:p\in 
\Gamma_{{{{\mathbb P}^\theta_\lambda}}}\}\mbox{ ''.}$$ Note that $\dot{U}$ is (a name for) a subset of $\lambda^+$. Let $\dot{F}$ be a ${{{\mathbb P}^\theta_\lambda}}$–name such that $${\Vdash}_{{{{\mathbb P}^\theta_\lambda}}}\mbox{`` }\dot{F}=\{f\in 2^{\textstyle\dot{U}}: (\forall
p\in\Gamma_{{{{\mathbb P}^\theta_\lambda}}})(f\restriction u^p\in\dot{F}^p)\}\mbox{ ''.}$$

\[3.4\] Assume $\theta<\lambda$ are regular, $\lambda^{<\lambda}=\lambda$. Then in ${{\bf V}}^{{{{\mathbb P}^\theta_\lambda}}}$:

1.  $\dot{F}$ is a non-empty closed subset of $2^{\textstyle\dot{U}}$, and ${\dot{{{\mathbb B}}}^\theta_\lambda}$ is the Boolean algebra generated ${{\mathbb B}}_{(\dot{U},\dot{F})}$ (see Definition \[0.C\]);

2.  $|\dot{U}|=|{\dot{{{\mathbb B}}}^\theta_\lambda}|=\lambda^+$;

3.  For every subalgebra ${{\mathbb B}}\subseteq{\dot{{{\mathbb B}}}^\theta_\lambda}$ of size $\lambda^+$ we have ${{\rm Depth}}^+({{\mathbb B}})>\theta$.

2)Note that if $p\in{{{\mathbb P}^\theta_\lambda}}$, $\sup(u^p)<j<\lambda^+$ then there is a condition $q\geq p$ such that $j\in u^q$. 