Consider the following bicomplemented lattice L of the form $ L=(L, \wedge, \vee, \neg, D, 0, 1)$, where the base set, is order-isomorphic to an ordinal. Furthermore, $\neg x = max \{y: x \wedge y = 0 \}$ exists for every $x \in L$ as well as $Dx = min \{y \in L: x \vee y =1 \}$ exists for every $x \in L$. Notice that we are only considering linear L. Now we can define a negation $\sim x =_{df} Dx \wedge (x \vee \neg x)$ and an implication $ x\leadsto y =_{df} \neg(x \wedge \neg y)$, that we will interpret respectively as negation and implication in our L-valued models.
Now we can define by transfinite recursion a L-valued model of set theory $V^{L}$. Take the following particular case $L_4$= $(\{1,b,a,0\}, \wedge, \vee, \neg, D, 0, 1)$ where $1 \geq b \geq a \geq 0$. Then we define the corresponding model $V^{L_4}$ as usual. Now, is it possible to define a sentence $\varphi$ in the language of set theory (without parameters ) such that $[\varphi]^{L}=a$ ? If you answer is negative, how would you prove it ? By induction on the complexity of sentences (so you take $\Pi_1$ and $\Sigma_1$ sentences as base case and then you consider more complex sentences for your step)?