if and only if }\quad (\forall\zeta<\gamma)(\xi
    \leq\zeta\ \Rightarrow\ h^p(i,\zeta)\geq\theta).$$

1)Should be clear (an easy induction).

2)Suppose that $p\in{{{\mathbb P}^\theta_\lambda}}$ and $j\in u^p$ are a counterexample with the minimal possible value of ${{\rm ht}}(p)$. Necessarily ${{\rm ht}}(p)$ is a limit ordinal, $p=\langle p_\xi:\xi<{{\rm ht}}(p)\rangle$, ${{\rm ht}}(p_\xi)=\xi$ and $\zeta<
\xi<{{\rm ht}}(p)\ \Rightarrow\ p_\zeta\leq_{\rm pr} p_\xi$. Let $\xi<{{\rm ht}}(p)$ be the first ordinal such that $j\in u^{p_\xi}$. By the choice of $p$, the set $\{\beta\leq\xi: h^p(j,\beta)<\theta\}$ is finite, but clearly $h^p(j,\beta)
\geq\theta$ for all $\beta\in (\xi,{{\rm ht}}(p))$.

3)An easy induction on ${{\rm ht}}(q)$ (with fixed $p$).

4)We show this by induction on ${{\rm ht}}(p)$. Suppose that ${{\rm ht}}(
p)=\alpha+1$, so $p=\langle\zeta^*,\tau^*,n^*,u^*,\langle p_\xi,v_\xi:\xi<
\theta\rangle\rangle$, and $i,j\in u^p$ are distinct. If $i,j\in u^{p_\xi}$ for some $\xi<\theta$, then by the inductive hypothesis we find $\beta<
\alpha$ such that $$\theta\neq h^p(i,\beta)=h^{p_\xi}(i,\beta)\neq h^{p_\xi}(j,\beta)=
h^p(j,\beta)\neq\theta.$$ If $i\in u^{p_\xi}\setminus u^*$, $j\in u^{p_\zeta}\setminus u^*$ and $\xi,
\zeta<\theta$ are distinct, then look at the definition of $h^p(i,\alpha)$, $h^p(j,\alpha)$ – these two values cannot be equal (and both are distinct from $\theta$). Finally suppose that ${{\rm ht}}(p)$ is limit, so $p=\langle
p_\xi:\xi<{{\rm ht}}(p)\rangle$. Take $\xi<{{\rm ht}}(p)$ such that $i,j\in u^{p_\xi}$ and apply the inductive hypothesis to $p_\xi$ getting $\beta<\xi$ such that $h^p(i,\beta)\neq h^p(j,\beta)$ 