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!