$\Pi_2$ formula true for $L_{{\omega_1}^L}$ but not for $L$ I tried the formula 
[${\forall x}{ \exists} {f} $$ {x}$ is an ordinal ${\land} {f}$is an injective function ${\land}$  $dom f =x$ ${\land}$ $ran f$ ${\subseteq} {\omega}$$]{ \lor}$ $x$ is not an ordinal. 
This formula works if I assume $V=L$, because then $\omega_1 ={\omega_1}^L$ and by using mostowski isomorphism, one can show that the formula holds. But without $V=L$, I don't know how to do that.
${\omega_1}^L$ denotes the first uncountable ordinal in $L$.
 A: You already have the right candidate and it now suffices to prove that
$$
L_{\omega_{1}^{L}} \models \text{every set is countable}.
$$
Let us work in $L$ and thus write $\omega_{1}$ for $\omega_{1}^{L}$, $H_{\theta}$ for $\left( H_{\theta} \right)^{L}$ and so on. Let $x \in L_{\omega_{1}}$. Then there is
some injective $f \colon x \to \omega$, $f \in L$. There are now two
ways to prove that such an $f$ exists in $L_{\omega_{1}}$:
Version 1: Note that $L_{\omega_{1}} = H_{\omega_{1}}$. Since
$H_{\omega_{1}} \prec_{1} V$ and $L \models V = L$, we obtain that
$L_{\omega_{1}} \models \text{there is an injective function } f
\colon x \to \omega$.
Version 2: Let $F$ be the transitive closure of $f$ and note that $F$
is countable. Fix some sufficiently large $\theta$ and let
$H \prec H_{\theta}$ be the Skolem hull of $\{ F , f \}$ in
$H_{\theta}$. Let $\sigma \colon T \equiv H$ be the transitive
collapse of $H$. Since $T$ is countable, the condensation lemma yields
that $T = L_{\alpha}$ for some countable ordinal $\alpha$. Now, since
$F$ is transitive and $f \subseteq F$, we know that the Mostowski
collapse restricted to $f$ is the identity and thus $f =
\sigma^{-1}(f) \in T = L_{\alpha} \subseteq L_{\omega_{1}}$. (This proves the stronger fact that any such $f$ actually lies in $L_{\omega_{1}}$.)
A: Two points:


*

*You need to exclude the case where $x$ is not an ordinal. It is true that every set in $L_{\omega_1^L}$ is countable there. 

*The assumption that $V=L$ is true in $L$, which is where you are interested in showing this sentence is false. 
