$q^0_\xi$ to $q^1_\xi$.

2.  If ${{\rm ht}}(q_0)={{\rm ht}}(q_1)=\alpha+1$, $\alpha<\lambda$, and $q_\ell=
    \langle\zeta^*_\ell,\tau^*_\ell,n^*_\ell,u^*_\ell,\langle q^\ell_\xi,
    v^\ell_\xi:\xi<\theta\rangle\rangle$ (for $\ell<2$), then $\zeta^*_0=
    \zeta^*_1$, $\tau^*_0=\tau^*_1$, $n^*_0=n^*_1$, $H{\restriction}u^{q_\xi^0}$ is an isomorphism from $q^0_\xi$ to $q^1_\xi$ and $H[v^0_\xi]=v^1_\xi$ (for $\xi<
    \theta$).

3.  $F^{q_0}=\{f{\circ}H:f\in F^{q_1}\}$.

4.  Assume $p_0\leq q_0$. Then there is a unique condition $p_1\leq q_1$ such that $H{\restriction}u^{p_0}$ is the isomorphism from $p_0$ to $p_1$.\

1), 2)Straightforward (for (1) use Lemma \[3.1\](7)).\
3), 4)Easy inductions on ${{\rm ht}}(q_0)$ using (1), (2) above.

\[defptran\] By induction on $\alpha<\lambda$, for conditions $p,q\in P^{\theta,
\lambda}_\alpha$ such that $p\leq^\alpha q$, we define [*the $p$–transformation $T_p(q)$ of $q$*]{}.

-   If $\alpha=0$ (so necessarily $p=q$) then $T_p(q)=p$.

-   Assume that ${{\rm ht}}(q)=\alpha+1$, $q=\langle\zeta^*,\tau^*,n^*,u^*,
    \langle q_\xi,v_\xi:\xi<\theta\rangle\rangle$.

    If $p\leq q_\xi$ for some $\xi<\theta$, then let $\xi^*$ be such that $p\leq
    q_{\xi^*}$. Next for $\xi<\theta$ let $q_\xi'=T_{H_{\xi^*, \xi}(p)}(q_\xi)$, where $H_{\xi^*,\xi}$ is the isomorphism from $q_{\xi^*}$ to $q_\xi$. Define $T_p(q)=\langle\xi^*,\tau^*,n^*,u^*,\langle q_\xi',v_\xi: \xi<\theta\rangle
    \rangle$.

    Suppose now that $p=\langle\zeta^{**},\tau^*,n^*,u^*,\langle p_\xi,v_\xi:\xi 
    <\theta\rangle\rangle$ and $u^{p_\xi}=u^{q_\xi}$, $p_\xi\leq q_\xi$ (for $\xi<\theta$). Let $q_\xi'=T_{p_\xi}(q_\xi)$ and put $T_p(q)=\langle
    \zeta^{**},\tau^*,n^*,u^*,\langle q_\xi',v_\xi:\xi<\theta\rangle\rangle$.

-   Assume now that ${{\rm ht}}(q)$ is a limit ordinal and $q=\langle q_\xi:\xi<
    {{\rm ht}}(q)\rangle$.

    If ${{\rm ht}}(p)<{{\rm ht}}(q)$ then $p\leq q_{\varepsilon}$ for some ${\varepsilon}<{{\rm ht}}(q)$, and we may choose $q_\xi'$ (for $\xi<{{\rm ht}}(q)$) 