Proving the schema of separation from replacement

Question

I have been looking at how separation can be deduced from replacement. I have found already this question: Proving Separation from Replacement

It was very enlightening, but it caused some thoughts to arise. First of all, I have noticed that people have different versions of the axiom of replacement. Omitting for brevity the requirement on $P$ to be a functional predicate, the version I use is the following: $\forall A \, \exists B \, \forall C, (C \in B \leftrightarrow \exists D \,(D \in A \wedge P(D,C)))$

I noticed other people had a version with $\exists! B$. The thing that puzzles me is that if you take "my" version of the axiom and instantiate B with $X$ and then with $Y$, it is very easy to prove that $X=Y$ (without even using any set theory, just logic). And this means to me that there is no reason to have $\exists! B$ as an axiom, because it can be very easily derived. My first question would be: am I right or am I missing something?

Moving on, I seem to see two different proofs for the separation schema and one seems massively easier than the other one, so yet again I am puzzled. The easy one is proposed by Mauro in the page I linked. All he does is choosing an functional predicate $\phi(u) \land u = v$ to be used in the replacement axiom, where $\phi$ is the predicate we want to apply separation to. Then, everything follows relatively naturally.

I also see another more complicated way of doing this that I really do not understand. I see two different cases being tackled. The first case is when there actually exist values of $u$ in the starting set (say $A$) that satisfy $\phi$ and the second case is when they do not exist. Honestly, in Mauro's proof, I can't see a place where things don't work if you choose a $\phi$ that is not true for any $u \in A$.

The complex proof then uses a functional predicate that is $(\phi(u) \land v=u) \vee ((\neg\phi(u)) \land v=\bar{v})$, where $\bar{v}$ is one of the values for which $\phi$ holds.

I suppose this proof is well-known but the place where I found it is a youtube video: https://www.youtube.com/watch?v=AAJB9l-HAZs&t=3757s You can find it around minute 1:06:19.

My question is whether Mauro's proof is all I need or there is a reason that escapes me for the more complicated proof.

Answer 1

As for your first question, you are correct: there is no need to assert uniqueness of $B$ in the statement of replacement, and in fact your version without uniqueness is the more usual way to state it. However, you do need some set theory, not just "logic": you need the axiom of extensionality. Without extensionality, you could have two different sets which have the same elements.

As for the two different proofs of separation, they are a result of using two different definitions of "functional predicate", and thus two different (though equivalent) statements of replacement. It appears that you are defining $F(u,v)$ to be a functional predicate if for every $u$, there is at most one $v$ such that $F(u,v)$. However, it is also common to define $F(u,v)$ to be a functional predicate if for every $u$, there is exactly one $v$ such that $F(u,v)$. In that case, your predicate $\phi(u)\wedge u=v$ is not functional (unless $\phi(u)$ is always true), and so you need to modify it before using replacement.