I have seen two slightly different forms of the axiom schema of specification.
a) For each formula $\psi(x,t)$, the following is an axiom:$\forall t\forall A\exists B\forall x(x\in B\leftrightarrow x\in A\land \psi(x,t))$.
b) For each formula $\psi(x)$, the following is an axiom:$\forall A\exists B\forall x(x\in B\leftrightarrow x\in A\land\psi(x))$.
I know that a) implies b) because a formula $\psi(x)$ in which $t$ occurs as a bound variable can also be written as $\psi(x,t)$. And my question is, does b) imply a)?
And I found that one could use either a) or b) to construct the set $\{x\in\mathbb{C}:x\in\mathbb{R}\}$. a) provides an easy approach; using b), however, requires lots of effort.
Using a): Let $\psi(x,t)=x\in t$. Then we have $\forall t\forall A\exists B\forall x(x\in B\leftrightarrow x\in A\land x\in t)$. In particular, for sets $\mathbb{R}$ (playing the role of $t$) and $\mathbb{C}$ (playing the role of $A$), there exists a (unique) set $B$ such that $\forall x(x\in B\leftrightarrow x\in\mathbb{C}\land x\in\mathbb{R})$.
Using b): We need to first rewrite the property $x\in\mathbb{R}$ as a formula $\psi(x)$. It is this process that requires lots of effort. After obtaining the formula $\psi(x)$, one could deduce from b) that $\forall A\exists B\forall x(x\in B\leftrightarrow x\in A\land\psi(x))$.In particular, for the set $\mathbb{C}$ (playing the role of $A$), there exists a (unique) set $B$ such that $\forall x(x\in B\leftrightarrow x\in \mathbb{C}\land\psi(x))$, i.e., $\forall x(x\in B\leftrightarrow x\in \mathbb{C}\land x\in\mathbb{R})$.
I'm so confused about these two approaches that seem to be so different from each other.
A related question: One can easily justify the existence of the intersection of any two sets using a). Just take $\psi(x,t)=x\in t$ and then deducing that $\forall t\forall A\exists B\forall x(x\in B\leftrightarrow x\in A\land x\in t)$.
But is it possible to construct $A\cap B$ for all $A$ and $B$ using b)? Here is what puzzles me:
Suppose that $A$ and $B$ are any sets. If we intend to apply b) to construct the intersection $A\cap B$, then a formula $\psi(x)$ is needed.It seems that "$x\in B$" is the right formula. However, as I understand it, $B$ is a free variable in "$x\in B$", while the symbol $\psi(x)$ indicates that the only possibly free variable is $x$. I can't tell whether $B$ is free or not in "$x\in B$". If not, then it is a $\psi(x)$ and we're finished.
