TITLE: Namba forcing and semiproperness
QUESTION [17 upvotes]: This question is the result of leaving "Proper and Improper Forcing" on my nightstand by accident.
Is the statement "Namba forcing is semiproper" known to be equiconsistent with some more standard large cardinal axiom?
I know the statement can be forced assuming a measurable cardinal, and it implies Chang's Conjecture (hence the existence of $0^\sharp$), and the Weak Reflection Principle (WRP) at $\omega_2$. Are narrower bounds known?

REPLY [4 votes]: Here is a direct construction of semigeneric conditions for Namba forcing (actually for Namba$^{\prime}$ for convenience) using $SCC$, without using games.
Let $\theta$ be large enough and regular, $X\prec H_\theta$ countable and $p\in X\cap Nm^\prime$.   We construct a descending sequence $\langle p_n\mid n<\omega\rangle$ of condtions and make sure that the $n$th level remains constant after $n$ steps, so that we get a fusion $p_\omega$ in the end. 
Let $\langle \dot \xi_n\mid n<\omega\rangle $ be an enumeration of all names for countable ordinals in $X$. Put $p_0=p$ and assume $p_n$ is constructed. Let $\eta$ be on the $n$th level of $p_n$. Find $r_\eta\leq p_n\upharpoonright \eta$ that has stem $\eta$ and decides as many of the $\dot\xi_0,\dots, \dot\xi_n$ as possible and let $\zeta_\eta$ be the supremum of the relevant decisions. We get $p_{n+1}$ by replacing all such subtrees $p_n\upharpoonright \eta$ by the corresponding $r_\eta$.
By making sure that all our choices are minimal with respect to some wellorder in $X$, we can make sure that all $p_n$ are in $X$ (even though $p_\omega$ is not). Even more, whenever $X\subseteq Y\prec H_\theta$ and $\eta\in p_\omega\cap Y$ then $\zeta_\eta\in Y$. 
Now we will thin out $p_\omega$ (we need $SCC$ to make sure that we do not cut off too many branches).  Let 
$$q=\{\eta\in p_\omega\mid \exists Y\text{ countable with }  X\sqsubseteq Y\prec H_\theta\wedge \eta\in Y\}$$
I claim that $q$ is semigeneric for $X$. First we have to see that $q$ is a condition. Suppose that $\eta\in q$ and that $Y$ witnesses this. Let $n=lh(\eta)$. Note that the direct successors of $\eta$ in $p_\omega$ are the same as the ones in $p_{n+1}\in Y$, so that we can find an enumeration $(\mu_\alpha)_{\alpha<\omega_2}\in Y$ of them. By $SSC$, there are unboundedly many $\beta<\omega_2$ for which there are countable $Y\sqsubseteq Z\prec H_\theta$ with $\beta\in Z$ an thus $\mu_\beta\in Z$. Hence every point in $q$ has $\omega_2$-many successors. 
Finally suppose that $r\leq q$ decides $\dot \xi_n$ as $\xi$. Let $\eta$ be the stem of $r$, we may assume that $lh(\eta)\geq n$. By the construction of $p_{lh(\eta)+1}$, $\xi\leq\zeta_\eta$. Now let $Y$ witness that $\eta\in q$. We thus have $\xi\leq\zeta_\eta\in Y\cap\omega_1=X\cap\omega_1$ so that $\xi\in X$, as desired.