If $\psi(x,\ldots)$ is a $\Delta_0$-formula then $\neg\psi(x,\ldots)$ if and only if $||\psi(\check{x},\ldots)||=0$ Come from Jech's Set Theory.
Exercise 14.13
(i) If $x=y$ then $||\check{x}=\check{y}||=1$ and if $x\neq y$ then $||\check{x}=\check{y}||=0$.
(ii)  If $x\in y$ then $||\check{x}\in \check{y}||=1$ and if $x\notin y$ then $||\check{x}\in \check{y}||=0$.
I use this exercise as a crucial step in Lemma 14.21, and so I think this exercise suggests that 

if $\psi(x,\ldots)$ is a $\Delta_0$-formula then 
(i) If $\psi(x,\ldots)$ holds then $||\psi(\check{x},\ldots)||=1$ and
(ii) If $\neg\psi(x,\ldots)$ holds then$||\psi(\check{x},\ldots)||=0$.

In fact, I have proved it. But I'm still not sure whether this proposition is true since Jech didn't mention it.
Edit. I still have some trouble in Lemma 14.21. I want to prove that  

If $\psi(x,\ldots)$ is a $\Delta_0$-formula then $\psi(x,\ldots)$ if and only if $||\psi(\check{x},\ldots)||=1.$

My approach:
Prove by induction on the complexity of $\varphi$. Suppose $\varphi$ is of the form $(\exists x\in\check{y})\psi(x,\ldots)$.
$
\begin{array}{rcl}
||(\exists x\in\check{y})\psi(x,\ldots)||=1&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}\check{y}(\check{x})\cdot\||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&\displaystyle\sum_{x\in y}||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&(\exists x\in y)||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&(\exists x\in y)\psi(x,\ldots)
\end{array}
$
But it seems that $\displaystyle\sum_{x\in y}||\psi(\check{x},\ldots)||=1$ doesn't imply that $(\exists x\in y)||\psi(\check{x},\ldots)||=1$ in general, and I don't know the value of $||\psi(\check{x},\ldots)||$ since it is possible that $\psi$ is consistent relative to ZFC.
The problem happens in the proof for $\psi$ is an atomic formula as well.
 A: OK I think the followings work, and I found something weird.
The proposiion

(*) If $\psi(x,\ldots)$ is a $\Delta_0$-formula then $\psi(x,\ldots)$ if and only if $||\psi(\check{x},\ldots)||=1$

is equivalent to

(**) If $\psi(x,\ldots)$ is a $\Delta_0$-formula then $\neg\psi(x,\ldots)$ if and only if $||\psi(\check{x},\ldots)||=0,$

so we only need to prove one of them.
But when we use induction on the complexity of $\psi$, these two propositions are different since $\neg\psi$ has more complexity than $\psi$; in this condition, we cannot derive (**) from (*), and vice versa.
Moreover, if we assume (**) in the inductive step, it gives more information than (*), which is weird:

$
\begin{array}{rcl}
||(\exists x\in\check{y})\psi(x,\ldots)||=1&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}\check{y}(\check{x})\cdot\||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&\displaystyle\sum_{x\in y}||\psi(\check{x},\ldots)||=1\\
&\color{red}{\leftarrow}&(\exists x\in y)||\psi(\check{x},\ldots)||=1\\
&\leftrightarrow&(\exists x\in y)\psi(x,\ldots)
\end{array}
$

and

$
\begin{array}{rcl}
||(\exists x\in\check{y})\psi(x,\ldots)||=0&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}\check{y}(\check{x})\cdot\||\psi(\check{x},\ldots)||=0\\
&\leftrightarrow&\displaystyle\sum_{\check{x}\in\textrm{dom}(\check{y})}||\psi(\check{x},\ldots)||=0\\
&\leftrightarrow&\displaystyle\sum_{x\in y}||\psi(\check{x},\ldots)||=0\\
&\leftrightarrow&(\forall x\in y)||\psi(\check{x},\ldots)||=0\\
&\leftrightarrow&(\forall x\in y)\neg\psi(x,\ldots).
\end{array}
$

So it is easier to prove $(**)$ directly, and eventually we'll have (*) as well.
Edit.
Well, it doesn't work for $\psi$ is $x=y$.
