{{\rm otp}}(U[p,Z_1])$, and the order preserving isomorphism $\pi:U[p,Z_0]\longrightarrow U[p,Z_1]$ satisfies

1.  $(\forall\ell<k)(h^p(i,\alpha^0_\ell)=h^p(\pi(i),
    \alpha^1_\ell))$,

    where $\{\alpha^x_0,\ldots,\alpha^x_{k-1}\}$ is the increasing enumeration of $Z_x$ (for $x=0,1$).

We prove this by induction on $|Z_0|=|Z_1|$ (for all $p,Z_0,Z_1$ satisfying the assumptions).

[Step]{} $|Z_0|=|Z_1|=1$; $Z_0=\{\alpha^0_0\}$, $Z_1=\{\alpha^1_0\}$.\
Take the $\alpha^x_0+1$–component $q_x$ of $p$ such that $q_x\leq_{\rm pr}
p$. Then, for $x=0,1$, $q_x=\langle\zeta,\tau,n, u^x,\langle q^x_\xi,
v^x_\xi:\xi<\theta\rangle\rangle$, and for each $i\in v^x_\xi$, $\beta<
\alpha^x_0$ we have $h^{q^x_\xi}(i,\beta)\geq\theta$. Also, if $i^x_0=
\min(u^{q^x_\zeta}\setminus u^x)$ and $\beta<\alpha^x_0$, then $h^{q^x_\zeta}(i^x_0,\beta)\geq\theta$. Consequently, $n=|v^x_\xi|\leq 1$, and if $n=1$ then $\{i^x_0\}=v^x_\zeta$ (remember Lemma \[3.1\](4)). Moreover, $$U[p,Z_x]=U[q_x,Z_x]=\{H^x_{\xi,\zeta}(i^x_0):\xi<\theta\},$$ where $H^x_{\xi,\zeta}$ is the isomorphism from $q^x_\zeta$ to $q^x_\xi$. Now it should be clear that the mapping $\pi:H^0_{\xi,\zeta}(
i^0_0)\mapsto H^1_{\xi,\zeta}(i^1_0):U[p,Z_0]\longrightarrow U[p,Z_1]$ is the order preserving isomorphism (remember clause $(\gamma)$ of the definition of $P^*_{\alpha+1}$), and it has the property described in $(\otimes)$.

[Step]{} $|Z_0|=|Z_1|=k+1$; $Z_0=\{\alpha^0_0,\ldots,\alpha^0_k
\}$, $Z_1=\{\alpha^1_0,\ldots,\alpha^1_k\}$.\
Let $$\Upsilon_p(Z_0)=\Upsilon_p(Z_1)=\langle\zeta_\ell,\tau_\ell,n_\ell,\langle
g_\ell,h^\ell_0,\ldots,h^\ell_{n_\ell-1}\rangle:\ell\leq k\rangle.$$ For $x=0,1$, let $q_x=\langle\zeta,\tau,n, u^x,\langle q^x_\xi,v^x_\xi:\xi<
\theta\rangle\rangle$ be the $\alpha^x_k+1$–component of $p$ such that $q_x
\leq_{\rm pr} p$. The sets $Z_x\cap\alpha^x_k$ (for $x=0,1$) are $q^x_\xi$–closed for every $\xi<\theta$, and clearly $\Upsilon_p(Z_0\cap
\alpha^0_k)= \Upsilon_p(Z_1\cap\alpha^1_k)$. Hence, by the inductive hypothesis, ${{\rm otp}}(U[q^0_\xi,Z_0\setminus\{\alpha^0_k\}])={{\rm otp}}(U[q^1_\xi,Z_1
\setminus\{\alpha^1_k\}])$ (for each $\xi<\theta$), and the order preserving mappings $\pi_\xi:U[q^0_\xi,Z_0\setminus\{ \alpha^0_k\}]\longrightarrow
U[q^1_\xi, Z_1\setminus\{\alpha^1_k\}]$ satisfy the demand in $(\otimes)$. 