At the start of "Suitable Extender Models I" Hugh Woodin makes the following statement.


"Suppose there is a proper class of Woodin cardinals. Then for each $\Sigma\_2$-sentence, $\phi$, the $\Sigma\_2$-sentence, $\phi^{\*}$, which 
asserts that $\phi$ holds in some forcing extension of $V$ , is absolute
between $V$ and all generic extensions of $V$."


This doesn't seem to be proved in Paul Larson's book "The Stationary Tower". Is there a reference for this result?



Seeking reference for absoluteness result in Omega-logic
We suppose there is a proper class of Woodin cardinals. We claim that the statement $\phi^\*$ which says that ``$\exists \mathbb{B} \exists\alpha(\mathbb{B}\text{ is a complete Boolean algebra and } \alpha\text{ is an ordinal}\wedge V\_{\alpha}^{\mathbb{B}}\models \phi)"$, where $\phi$ is a $\Sigma\_2$ sentence is absolute between $V$ and $V^{\mathbb{H}}$ for any complete Boolean algebra $\mathbb{H}$.


Notice that $\phi^\*$ is a $\Sigma\_2$ sentence since the statement $``V^{\mathbb{B}}\models \phi"$ is $\Delta\_2$ and the statement $``\mathbb{B}$ is a complete Boolean algebra" is just $\Pi\_1$. Let then $\phi$ be any $\Sigma\_2$-sentence.


First, let $\mathbb{H}$ be any complete Boolean algebra. Let $\check{\gamma}\in V^{\mathbb{H}}$. We want to find a complete Boolean algebra $\mathbb{B}\in V^{\mathbb{H}}$ such that we have $$V^{\mathbb{H}}\models ``V\_{\check{\gamma}}^{\mathbb{B}}\models \phi".$$ But since the forcing extension of a $V\_{\gamma}$ is the same as the $V\_{\gamma}$ of a forcing extension, that is $V\_{\gamma}^{\mathbb{H}}=(V\_{\check{\gamma}})^{V^{\mathbb{H}}}$. Since there is a complete Boolean algebra $\mathbb{B}$ in $V$ such that $\phi$ holds in the extension $V^{\mathbb{B}}$, one may then find a complete Boolean algebra $\mathbb{B}\in V^{\mathbb{H}}$ such that we can conclude that $$V^{\mathbb{H}}\models ``V^{\mathbb{B}}\_{\check{\gamma}}\models \phi".$$


Next assume that for any complete Boolean algebra $\mathbb{H}$, the statement $\phi^\*$ hold in $V^{\mathbb{H}}$ for some $\Sigma\_2$ sentence $\phi$. Let $\mathbb{H}$ have cardinality $\gamma$. By assumption we let $\gamma<\delta$ be a Woodin cardinal. The forcing notion which will force $\phi$ over $V$ is given by the stationary tower forcing. Define the set 


$$Y=\{X: X\prec H(\gamma^+)\wedge \vert X\vert =\aleph\_0\}.$$


Define the stationary tower forcing to be the forcing $$\mathbb{Q}\_{<\delta}=\{X: X\text{ is stationary and }X\in V\_{\delta}\}.$$ The order is defined by $X\leq Y$ if and only if $\bigcup Y\subseteq \bigcup X$ and $\{Z\cap (\bigcup Y): Z\in X\}\subseteq Y$. We must have that $Y\in \mathbb{Q}\_{<\delta}$ since if $F: H(\gamma^+)^{<\omega}\to H(\gamma^+)$ and building the elementary substructure of $H(\gamma^+)$ closed under $F$, let's call it $X$, we have $$X\prec H(\gamma^+)$$ and $$\vert X\vert =\aleph\_0$$ and $F"X\subseteq X$. This is exactly what's required by stationarity. Let $G\subseteq \mathbb{Q}\_{<\delta}^X$ be $V$-generic where $\mathbb{Q}\_{<\delta}^X$ is the stationary tower forcing restricted to $X$. BY the properties of the stationary tower forcing we can find in $V[G]$ an elementary embedding $$j: V[G] \to M$$ where $M$ is a transitive model of $\text{ZFC}$ and such that, given that $\delta$ is a Woodin cardinal $$M^{<\delta}\subseteq M$$ and $H(\kappa^+)$ is countable in $V[G]$. Thus we have that $\mathbb{H} \in M$. We may then find a generic filter in $M$ for $\mathbb{H}$. Call this generic $H$ and we thus have $$V[H]\subseteq V[G].$$ Looking at the quotient forcing, choose a forcing notion $\mathbb{T}$ and a generic $K$ for $\mathbb{T}$ such that $$V[H][K]=V[G]$$ Finally we conclude that $V[G]\models \phi$ because $V[G]=V[H][K]$ and $V[H][K]\models \phi$ by assumption since $M\models \phi$


