The assertion $$\varphi=\forall x(x\in L\land\mathrm{rank}(x)=\alpha\rightarrow x\in L_{\alpha+1})$$ is not provable in $\mathsf{ZFC}$: under $V=L$ there's uncountably many sets of rank $\omega$ in $L$, but $L_{\omega+1}$ is countable. Is $\neg\varphi$ provable in $\mathsf{ZFC}$ or is $\varphi$ consistent with $\mathsf{ZFC}$?
I'm guessing that $\varphi$ is provably false in $\mathsf{ZFC}$, it would imply that $L$ is "a lot smaller" than $V$ and it seems like there should be counterexamples, but I wasn't able to find one. I tried arguing that $L$ is an inner model of $\mathsf{ZFC}$, so $\omega^L=\omega^V$, $\mathcal P(\omega)^L=(\mathcal P(\omega)^V\cap L)\in L$ and $L$ believes $\mathcal P(\omega)^L$ to be uncountable, but "being uncountable" is a $\Pi_1$-formula, hence not necessarily upward absolute between transitive classes, so I cannot conclude that $V$ also believes $\mathcal P(\omega)^L$ to be uncountable.
Well, clearly $\varphi$ is false. $\sf ZF$, in fact, already proves that.
The reason, of course, is that both the von Neumann hierarchy and the constructible hierarchy are absolute (just cardinality itself is not). So a set which has rank $\alpha$ in $V$ will have the same rank in $L$, and vice versa.
You can play a game with yourself, find more proofs why this $\varphi$ is provably false. I can think of at least one more.