h x\in \dot{ \mathcal{X}}$. It is clear that $\mathcal{X}$ is a stationary subset of $[\kappa]^\omega$ because $s_0$ forces that $\mathcal{X}$ meets every cub. Now apply **Axiom R** to choose $Y\in \mathcal{C}$ so that $\mathcal{X}\cap [Y]^\omega$ is a stationary subset of $Y$.

Now we obtain a contradiction (and thus a proof) by showing that there is an extension $s\in S$ of $s_0$ that forces that $\dot{ \mathcal{X}}\cap [Y]^\omega$ is stationary. Let $\{ y_\alpha :\alpha\in \omega_1\}$ be an enumeration of $Y$. Let $\mathcal{E}$ be the set of $\delta\in \omega_1$ such that $x_\delta = \{y_\alpha :\alpha\in \delta\}\in \mathcal{X}$. Notice that $\{ \{ y_\alpha : \alpha \in \delta \} : \delta\in \omega_1\}$ is a cub in $[Y]^\omega$. Thus it follows that $\mathcal{E}$ is stationary. In fact, if $\mathcal{E}'$ is any stationary subset of $\mathcal{E}$, then $\mathcal{E}'$ is also a stationary subset of $[Y]^\omega$.

For each $\delta\in \mathcal{E}$ choose $s_\delta\in S$ above $s_0$ so that $s_\delta\Vdash x_\delta\in \dot{ \mathcal{X}}$ (as per the definition of $ \mathcal{X}$). Now we have a name $\dot{ \mathcal{E}} = \{ (x_\delta, s_\delta) :
\delta\in \omega_1\}$. We prove that there is some $s\in S$ above $s_0$ that forces that $\dot{ \mathcal{E}}$ is stationary. Thus such an $s$ forces that $\dot{ \mathcal{X}}\cap [Y]^\omega$ is stationary as required.

Let $s_0$ be on level $\alpha_0$ of $S$. There is a $\gamma>\alpha_0$ so that each member of $S_\gamma$ decides if $\dot{ \mathcal{E}}$ is stationary. Also, for each $\bar s\in S_\gamma$ that forces $\dot{ \mathcal{E}}$ is not stationary, there is a cub $\mathcal{C}_{\bar
   s}$ of $\omega_1$ that $\bar s$ forces is disjoint from $\dot{ \mathcal{E}}$. Choose any $\delta $ in the intersection of those countably many cubs that is also in $\mathcal{E}$. Clearly if $\bar s\in S_\gamma$ is compatible with $s_\delta$, then $\mathcal{C}_{\bar s}$ did not exist since $\bar s \cap s_\delta$ would force that $\delta \in
 \mathcal{C}_{\bar s}\cap  \dot{ \mathcal{E}}$. T