such that ${{\rm ht}}(q_\xi')=\xi$, $\xi<
    \xi'<{{\rm ht}}(q)\ \Rightarrow\ q_\xi'\leq_{\rm pr} q_{\xi'}'$, and $q_\zeta'
    =T_p(q_\zeta)$ for $\zeta\in [{\varepsilon},{{\rm ht}}(q))$. Next we let $T_p(q)=\langle
    q_\zeta':\zeta<\theta\rangle$.

    If ${{\rm ht}}(p)={{\rm ht}}(q)$, $p=\langle p_\xi:\xi<{{\rm ht}}(p)\rangle$ and $p_\xi\leq
    q_\xi$ for $\xi>\delta$ (for some $\delta<{{\rm ht}}(p)$) then we define $T_p(q)=p$.

To show that the definition of $T_p(q)$ is correct one proves inductively (parallely to the definition of the $p$–transformation of $q$) the following facts.

\[3y1\] Assume $p,q\in{{{\mathbb P}^\theta_\lambda}}$, $p\le q$. Then:

1.  $T_p(q)\in{{{\mathbb P}^\theta_\lambda}}$, $u^{T_p(q)}=u^q$, ${{\rm ht}}(T_p(q))={{\rm ht}}(q)$,

2.  $p\leq_{\rm pr} T_p(q)\leq q\leq T_p(q)$,

3.  ${{\rm ht}}(p)={{\rm ht}}(q)\ \Rightarrow\ T_p(q)=p$,

4.  if $q'\in{{{\mathbb P}^\theta_\lambda}}$ is isomorphic to $q$ and $H:u^q\longrightarrow u^{q'}$ is the isomorphism from $q$ to $q'$, then $H$ is the isomorphism from $T_p(q)$ to $T_{H(p)}(q')$,

5.  if $q\leq_{\rm pr}q'$ then $T_p(q)\leq_{\rm pr} T_p(q')$.

\[3y2\] Every $\leq_{\rm pr}$–increasing chain in ${{{\mathbb P}^\theta_\lambda}}$ of length $<\lambda$ has a $\leq_{\rm pr}$–upper bound, that is the partial order $({{{\mathbb P}^\theta_\lambda}},\leq_{\rm
pr})$ is $(<\lambda)$–closed.

Let us recall that a forcing notion $({{\mathbb Q}},\leq)$ is [*$({<}\lambda)$–strategically closed*]{} if the second player has a winning strategy in the following game $\Game_\lambda({{\mathbb Q}})$.

The game $\Game_\lambda({{\mathbb Q}})$ lasts $\lambda$ moves. The first player starts with choosing a condition $p^*\in{{\mathbb Q}}$. Later, in her 