232 Logic as a Tool Proposition 185 For all sets A and B , A = B if and only if A ⊆ B and B ⊆ A Proof We have to prove an equivalence: “ if and only if ” We that by proving the implications in both directions To prove the implication “If A = B , then A ⊆ B and B ⊆ A” we assume the premise “A = B ” and try to derive the conclusion “A ⊆ B and B ⊆ A.” This is a conjunction, so we have to prove both “A ⊆ B ” and “B ⊆ A.” To prove that A ⊆ B , by definition of set inclusion we must show that every element of A is an element of B Let x be any (arbitrary) element of A Then, because A = B , we also have x ∈ B Since x ∈ A was arbitrary, we conclude that A ⊆ B Likewise, to show that B ⊆ A, let x be any element of B Because A = B , it follows that x ∈ A Since x ∈ B was arbitrary, we therefore conclude that B ⊆ A To prove “If A ⊆ B and B ⊆ A, then A = B ,” we assume the premise “A ⊆ B and B ⊆ A” and try to deduce A = B By the axiom EXT, we have to show that A and B have exactly the same elements Let x be an arbitrary element If x ∈ A, then since A ⊆ B , we have x ∈ B Similarly, if x ∈ B , then since B ⊆ A, we have x ∈ A We have therefore shown that, for any x, it holds that x ∈ A iff x ∈ B , that is, that A and B have the same elements, hence A = B , by EXT Note that the proof above is almost formalized in Natural Deduction As an exercise, formalize it completely in ND or in any deductive system of your choice for first-order logic for LZF This can be done in two ways: without explicitly using the symbol ⊆ or using ⊆ and adding its defining equivalence as an assumption: ∀x∀y (x ⊆ y ↔ ∀z (z ∈ x → z ∈ y )) Most of the other axioms of ZF are just statements of existence of some special sets or can be obtained by applying basic operations to already existing sets The empty set axiom states the existence of an “empty set”: EMPTY: ∃x∀y (¬y ∈ x) As an exercise, using EXT prove that there is only one empty set, denoted ∅ The pair set axiom states the existence of a set with elements being a given pair of sets: PAIR: ∀x∀y ∃z ∀u(u ∈ z ↔ u = x ∨ u = y ) As an exercise, using EXT prove that for any sets x, y there is only one pair set for them, denoted {x, y } The union set axiom states the existence of the union of any pair of sets: UNION: ∀x∀y ∃z ∀u(u ∈ z ↔ u ∈ x ∨ u ∈ y ) As an exercise, using EXT prove that for any sets x, y there is only one union set for them, denoted x ∪ y The powerset axiom states the existence of the set of all subsets of any given set: POWER: ∀x∃y ∀z (z ∈ y ↔ z ⊆ x)