(and both are not $\theta$).

5)Again, it goes by induction on ${{\rm ht}}(p)$. First consider a limit stage, and suppose that ${{\rm ht}}(p)=\gamma$ is a limit ordinal, $X\in[\gamma]^{\textstyle{<}\omega}$ and $p=\langle p_\xi:\xi<\gamma
\rangle$. Let $\xi<\gamma$ be such that $X\subseteq\xi$. By the inductive hypothesis we find $i\in u^{p_\xi}$ such that $\{\beta<\xi:h^p(i,\beta)<
\theta\}=X$. Applying clause (3) we may conclude that this $i$ is as required. Now consider a successor case ${{\rm ht}}(p)=\alpha+1$. Let $p=\langle
\zeta^*,\tau^*,n^*,u^*,\langle p_\xi,v_\xi:\xi<\theta\rangle\rangle$, and let $\xi<\theta$ be $\zeta^*$ if $\alpha\in X$, and be $\zeta^*+1$ otherwise. Apply the inductive hypothesis to $p_\xi$ and $X\cap \alpha$ to get suitable $i\in u^{p_\xi}$, and note that this $i$ works for $p$ and $X$ too.

6), 7)Straightforward.

\[defiso\] We say that conditions $p,q\in{{{\mathbb P}^\theta_\lambda}}$ are [*isomorphic*]{} if ${{\rm ht}}(p)=
{{\rm ht}}(q)$, ${{\rm otp}}(u^p)={{\rm otp}}(u^q)$, and if $H:u^p\longrightarrow u^q$ is the order isomorphism, then for every $\beta<{{\rm ht}}(p)$ $$(\forall j\in u^p)(h^p(j,\beta)=h^q(H(j),\beta)\ \&\ g^p(j,\beta)=g^p(H(j),
\beta)).$$ \[In this situation we may say that $H$ is the isomorphism from $p$ to $q$.\]

\[3x1\] Suppose that $q_0,q_1\in{{{\mathbb P}^\theta_\lambda}}$ are isomorphic conditions and $H$ is the isomorphism from $q_0$ to $q_1$.

1.  If ${{\rm ht}}(q_0)={{\rm ht}}(q_1)=\gamma$ is a limit ordinal, $q_\ell=\langle
    q^\ell_\xi: \xi<\gamma\rangle$ (for $\ell<2$), then $H{\restriction}u^{q_\xi^0}$ is an isomorphism from 