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!

  • 1
    From the way you are using ψ it seems to be a 1-place predicate, if so it can not be a contradiction, only a closed (e.g. fully quantified) formula can be that. So by "ψ is a contradiction" you probably mean ∃xψ(x) is a contradiction, i.e. ψ(x) is false for any value of x. Then formal proof will depend on the precise form of the definition of empty set, one definition is Ø:={x:x≠x}. But by assumption ψ(x) and x≠x have the same truth value on any x, so by the extensionality axiom {x:ψ(x)}=Ø.
    – Conifold
    Commented Sep 1, 2015 at 17:52

2 Answers 2


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).

  • Thanks Mauro. That's exactly what I was looking for. I had never seen the statement a∈{x:ψ(x)} ⇔ ψ(a), how come this isn't explicitly taken as an axiom in ZFC? Do you recommend any books that elaborate on this topic? Commented Sep 5, 2015 at 19:53
  • @EthanAlvaree - see e.g Patrick Suppes, Axiomatic set theory (1960 - Dover reprint), page 34 : it is not an axiom of ZFC but a general definition of an "operator" (the set-builder) used in any set theory, Commented Sep 5, 2015 at 20:12

@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.


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