$i^{\rm th}$ move, the first player chooses an open dense subset $D_i$ of ${{\mathbb Q}}$. The second player (in his $i^{\rm th}$ move) picks a condition $p_i\in{{\mathbb Q}}$ so that $p_0\geq
p^*$, $p_i\in D_i$ and $p_i\geq p_j$ for all $j<i$. The second player looses the play if for some $i<\lambda$ he has no legal move.

It should be clear that $({<}\lambda)$–strategically closed forcing notions do not add sequences of ordinals of length less than $\lambda$. The reader interested in this kind of properties of forcing notions and iterating them is referred to [@Sh:587], [@Sh:667].

\[3.3\] Assume that $\theta<\lambda$ are regular cardinals, $\lambda^{<\lambda}=
\lambda$. Then $({{{\mathbb P}^\theta_\lambda}},\leq)$ is a $(<\lambda)$–strategically closed $\lambda^+$–cc forcing notion.

It follows from Lemma \[3y1\](2) that if $D\subseteq{{{\mathbb P}^\theta_\lambda}}$ is an open dense set, $p\in{{{\mathbb P}^\theta_\lambda}}$, then there is a condition $q\in D$ such that $p\leq_{\rm
pr} q$. Therefore, to win the game $\Game_\lambda({{{\mathbb P}^\theta_\lambda}})$, the second player can play so that the conditions $p_i$ that he chooses are $\leq_{\rm 
pr}$–increasing, and thus there are no problems with finding $\leq_{\rm
pr}$–bounds (remember Proposition \[3y2\]).

Now, to show that ${{{\mathbb P}^\theta_\lambda}}$ is $\lambda^+$–cc, suppose that $\langle p_\delta:
\delta<\lambda^+\rangle$ is a sequence of distinct conditions from ${{{\mathbb P}^\theta_\lambda}}$. We 