Metamath Proof Explorer


Theorem elom

Description: Membership in omega. The left conjunct can be eliminated if we assume the Axiom of Infinity; see elom3 . (Contributed by NM, 15-May-1994)

Ref Expression
Assertion elom ( 𝐴 ∈ ω ↔ ( 𝐴 ∈ On ∧ ∀ 𝑥 ( Lim 𝑥 → 𝐴 ∈ 𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 eleq1 ⊢ ( 𝑦 = 𝐴 → ( 𝑦 ∈ 𝑥 ↔ 𝐴 ∈ 𝑥 ) )
2 1 imbi2d ⊢ ( 𝑦 = 𝐴 → ( ( Lim 𝑥 → 𝑦 ∈ 𝑥 ) ↔ ( Lim 𝑥 → 𝐴 ∈ 𝑥 ) ) )
3 2 albidv ⊢ ( 𝑦 = 𝐴 → ( ∀ 𝑥 ( Lim 𝑥 → 𝑦 ∈ 𝑥 ) ↔ ∀ 𝑥 ( Lim 𝑥 → 𝐴 ∈ 𝑥 ) ) )
4 df-om ⊢ ω = { 𝑦 ∈ On ∣ ∀ 𝑥 ( Lim 𝑥 → 𝑦 ∈ 𝑥 ) }
5 3 4 elrab2 ⊢ ( 𝐴 ∈ ω ↔ ( 𝐴 ∈ On ∧ ∀ 𝑥 ( Lim 𝑥 → 𝐴 ∈ 𝑥 ) ) )