# The empty set in Ackermann set theory

Let $$A$$ denote Ackermann set theory, as laid out e.g. here. It is a standard result due to Levy and Reinhardt that $$A$$ and $$ZF$$ are mutually interpretable in conservative extensions of one-another.

How do we see that the empty class is a set in $$A$$?

Despite scanning the relevant papers I seem unable to find a simple argument laying out how $$\emptyset\in\mathbb{V}$$. Naively, my thought was to consider the predicate $$\phi\equiv\ "x\ \text{is a set}\wedge x\neq x"$$ and take $$\emptyset$$ to be the set guaranteed by the reflection schema (axiom $$3$$ above) together with $$\phi$$, but I am concerned that by writing '$$x$$ is a set' in the definition of $$\phi$$ I am implicitly allowing $$\mathbb{V}$$ to appear, since this is equivalent to $$x\in\mathbb{V}$$ -- this would obviously prevent us from defining $$\emptyset$$ in this fashion (unless I am misunderstanding what it means for a constant symbol to 'be a parameter' in a predicate).

Does this approach work, or is some other trick required? Any tips are appreciated.

I believe you're overthinking the reflection scheme. Just use the formula $$F(x)\equiv x\not=x$$. Then:
• We trivially have $$\forall x(F(x)\rightarrow x\in V)$$, since no $$x$$ at all satisfies $$F$$.
• Consequently, reflection applies, and we get $$\{x:F(x)\}\in V$$ as desired.