Prove a logical formula is equivalent to the contradiction if and only if the set it describes is empty

Question

Let ψ be a well-formed-formula (wff). Prove that

(ψ ≡ ⊥) ⇔ {x:ψ(x)}=Ø

that is, the formula ψ is a contradiction if and only if the set it describes has no members.

Note This question is not about understanding why this is true. The idea seems intuitively simple: If ψ is a contradiction, then it is false and no element x in set theory can possibly satisfy it. That is, ψ(x) is false for any x. So the set {x:ψ(x)} has no members, i.e. is equal to the empty set Ø. The other direction is also not hard to understand (if {x:ψ(x)}=Ø, then no set x satisfies ψ, so ψ(x) must be false for any x, and so it makes sense that ψ is a contradiction.)

My question is, how could one prove this formally (that is, formalize my argument not in terms of loose English sentences like I have provided, but in strict formal logic?) Please note my knowledge of logic and structure is very limited apart from introductory set theory and the ZFC axioms. But I am trying to formalize my basic understanding of logic similar to how ZFC formalized set theory for me. Thanks for your help!

Answer 1

We have to remember that :

a ∈ {x:ψ(x)} ⇔ ψ(a)

(this is the definition of the "set-builder" symbol { _ : __ } ).

But {x:ψ(x)} = Ø, and thus :

for all a, a ∉ {x:ψ(x)} ⇔ for all a, ¬ψ(a).

Answer 2

@EthanAlvaree I assume that you consider psi an unary predicate. And that your statement psi equivalent false means: For all x: psi(x) is false.

If that's what you mean, then nearly by definition:

psi equivalent false if and only if The set of all x, which satisfy psi, is empty.