Proper class of Woodins and $\textsf{AD}_{\mathbb R}$-hypothesis The $\textsf{AD}_{\mathbb R}$-hypothesis is the statement that there is a $\lambda$ which is both a limit of Woodins and a limit of ${<}\lambda$-strongs. Are there any results relating the consistency of this statement to the consistency of a proper class of Woodins?
In this paper by Zeman he mentions that the $\textsf{AD}_{\mathbb R}$-hypothesis is the existence of a proper class of Woodins and strongs. To me that just sounds strictly stronger than the above-stated version - but that's perhaps just a change in terminology over time?
 A: The existence of a proper class of Woodin cardinals and a proper class of strong cardinals is strictly stronger in consistency strength over ZFC than the existence of a cardinal $\lambda$ that is a limit of Woodin cardinals and a limit of $<\lambda$-strong cardinals. The former implies the consistency of the latter and indeed the former implies that there are a proper class of such $\lambda$.
To see this, let $A$ be the class of Woodin cardinals and $B$ be the class of strong cardinals, and assume both of these are unbounded. Thus, both $A$ and $B$ have unboundedly many limit points. Since the limit points of an unbounded class are club and any two club classes intersect in a club, it follows that there are a proper class of $\lambda$ that are limits of Woodin cardinals and limits of strong cardinals. One can construct such a $\lambda$ explicitly as the limit of the following process: start with any Woodin cardinal, then take the first strong above it, the next Woodin above that, the next strong above that, and so on. Let $\lambda$ be the supremum of these choices. So $\lambda$ is a limit of Woodin Cardinals and strong cardinals, as large as desired. 
Thus, the assumption of a proper class of Woodin cardinals and a proper class of strong cardinals implies the consistency of the existence of a $\lambda$ that is a limit of Woodin cardinals and $<\lambda$-strong cardinals, simply by chopping off at an inaccessible cardinal. So the consistency strength is strictly stronger.
A: I think I got a solution. If you find any mistakes or if anything is unclear, please let me know!
Firstly, recall that the $\Omega>0$-hypothesis (or sometimes stated as $\theta_0<\Theta$-hypothesis), is the statement that there is a $\lambda$ which is the limit of Woodins and there is a $\kappa<\lambda$ which is ${<}\lambda$-strong. So this hypothesis is weaker than the $\textsf{AD}_{\mathbb R}$-hypothesis.

Proposition. The $\Omega>0$-hypothesis is consistency-wise stronger than a proper class of Woodins.

Proof. Assume $\lambda$ is a limit of Woodins and let $\kappa<\lambda$ be ${<}\lambda$-strong. Let $\delta<\lambda$ be the least Woodin above $\kappa$ and pick some $\gamma\in(\delta,\lambda)$. Then as $\kappa$ in particular is $\gamma$-strong, fix an elementary embedding $j_0\colon V\to\mathcal M_0$ such that $V_\gamma\subseteq\mathcal M_0$, $\text{crit }j=\kappa$ and $j(\kappa)>\gamma$. In $\mathcal M_0$ we still have that $\delta$ is Woodin, as the extenders witnessing Woodinness lie in $V_\delta$ and $V_\delta\subseteq V_\gamma\subseteq\mathcal M$.
We can then continue. At every successor stage $\alpha+1$ we use the $j_{0\alpha}(\gamma)$-strongness of $j_{0\alpha}(\kappa)$ to get a $j_{0\alpha}(\gamma)$-strong embedding $j_{\alpha+1}\colon\mathcal M_\alpha\to\mathcal M_{\alpha+1}$ and again note that $j_{0\xi}(\delta)$ is still Woodin in $\mathcal M_{\alpha+1}$ for every $\xi\leq\alpha$. At limit stages we take direct limits, with the same conclusion. We now have two cases.
Case 1. There is an ordinal $\alpha$ such that for every $\xi\geq\alpha$, $j_{0\xi}(\lambda)=j_{0\alpha}(\lambda)$.
In this case, after we get to stage $\alpha$, we iterate $j_{0\alpha}(\lambda)$ more times, which moves the extender on $j_{0\alpha}(\kappa)$ up to $j_{0\alpha}(\lambda)$. Then $V_{j_\xi(\lambda)}\cap \mathcal M_{\xi+j_\xi(\lambda)}$ satisfies $\textsf{ZFC}$ and it thinks that there is proper class of Woodins (as $j_\xi(\lambda)$ is an inaccessible limit of Woodins inside $\mathcal M_{\xi+j_\xi(\lambda)}$).
Case 2. For every ordinal $\alpha$ there is some $\xi>\alpha$ such that $j_{0\alpha}(\lambda)<j_{0\xi}(\lambda)$.
In this case we iterate $\text{On}$-many times, iterating $\lambda$ out of the universe and leaving a proper class of Woodins behind (namely, all the images of $\delta$). QED
EDIT: This result coupled with Joel's answer then gives us the following string of implications: Infinitely many Woodins < A proper class of Woodins $\leq$ $(\Omega >0)$-hypothesis $\leq$ $\textsf{AD}_{\mathbb R}$-hypothesis < A proper class of Woodins and a proper class of strongs.
