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](http://www.math.uci.edu/~mzeman/RESEARCH/ffs-0-120719.pdf) 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?



Proper class of Woodins and $\textsf{AD}_{\mathbb R}$-hypothesis
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.


