Here's [a very silly mistake I made recently](https://mathoverflow.net/questions/257613/getting-measures-especially-on-omega-2-from-potential-clubs): I claimed that if $\mathbb{P}\in L(\mathbb{R})$ is a forcing which adds a real, then $$(\*)\quad L(\mathbb{R})^{V^\mathbb{P}}=L(\mathbb{R})^\mathbb{P}.$$ While it is true (**EDIT: no it isn't, see the answer below**) that $L(\mathbb{R})^\mathbb{P}=L(\mathbb{R})^{(L(\mathbb{R})^\mathbb{P})}$, what I wrote above is nonsense: $V$ may have names for reals which are not in $\mathbb{R}$.


Indeed, from large cardinals we can prove that $(\*)$ *fails* - if it held, then examining $Col(\omega,\kappa)$ for $\kappa$ regular in $V$, we'd have that $L(\mathbb{R})$ contains a proper class of measurables; and this is known to not be the case, assuming large cardinals (specifically, from large cardinals there are no measurable cardinals in $L(\mathbb{R})$ above $\Theta$).


My question is around principles of the form $(\*)$. Specifically, for a definable class of forcings $\mathcal{C}$ which add a real, let (FC)$\_\mathcal{C}$ ("forcing conflation") be the statement $(\*)$ restricted to forcings in $\mathcal{C}\cap L(\mathbb{R})$. I'm interested in when (FC)$\_\mathcal{C}$ is compatible with large cardinals - specifically, a proper class of Woodins - and when it adds large cardinal strength to ZFC+proper class of Woodins. 


To make this concrete, I'll ask:



> 
> Is (FC)$\_{proper}$ consistent with ZFC + "There is a proper class of Woodins," relative perhaps to even stronger large cardinal hypotheses? If so, does ZFC + "There is a proper class of Woodins" + (FC)$\_{proper}$ have consistency strength beyond a proper class of Woodins?
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> 
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Forcing conflation for $L(\mathbb{R})$
$(FC)\_{\text{proper}}$ is false. If there are infinitely many Woodin cardinals with a measurable above, then after adding a single Cohen real, we do not have $L(\mathbb R)[c] = L(\mathbb R)^{V[c]}$. The reason is that $L(\mathbb R)[c]$ does not satisfy $\text{AD}$: in $L(\mathbb R)[c]$, the set of ground model reals is uncountable, but does not contain a perfect subset. (See Hamkins's answer to the question [Is there a perfect set of ground model reals in the Cohen extension?](https://mathoverflow.net/questions/251988/is-there-a-perfect-set-of-ground-model-reals-in-the-cohen-extension) which uses at worst countable choice.) On the other hand, $L(\mathbb R)^{V[c]}$ satisfies $\text{AD}$ by our large cardinal hypothesis. 


Note in this situation that $L(\mathbb R)^{V[c]}=L(\mathbb R)^{L(\mathbb R)[c]}\subsetneq L(\mathbb R)[c]$, since nice names for reals are all coded by reals. Therefore it is *not* true in general that $L(\mathbb R)^{L(\mathbb R)[r]} = L(\mathbb R)[r]$ (or that $L(\mathbb R)[r]\subseteq L(\mathbb R)^{V[r]}$) when $r$ is a generic real. The problem this time is that $\mathbb R$ (as computed in the ground model) need not be in $L(\mathbb R)^{L(\mathbb R)[r]}$.


The large cardinal assumption can actually be reduced to the assumption that $\text{AD}$ holds in $L(\mathbb R)$, because it can be proved under this hypothesis that $ L(\mathbb R)^{V[c]} = L(\mathbb R)^{L(\mathbb R)[c]}$ satisfies $\text{AD}$. You might be interested in Kechris and Woodin's paper Generic Codes, in which they prove this fact and analyze forcing over $L(\mathbb R)$ with Levy collapses, showing that for $\kappa<\delta^2\_1$ a reliable ordinal, if $G\subseteq \text{Col}(\omega, \kappa)$ is $L(\mathbb R)$-generic then in $L(\mathbb R)[G]$, there is a definable elementary embedding from $L(\mathbb R)$ into $L(\mathbb R)^{L(\mathbb R)[G]}$ 


