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 e. _om <-> ( A e. On /\ A. x ( Lim x -> A e. x ) ) )

Proof

Step Hyp Ref Expression
1 eleq1
 |-  ( y = A -> ( y e. x <-> A e. x ) )
2 1 imbi2d
 |-  ( y = A -> ( ( Lim x -> y e. x ) <-> ( Lim x -> A e. x ) ) )
3 2 albidv
 |-  ( y = A -> ( A. x ( Lim x -> y e. x ) <-> A. x ( Lim x -> A e. x ) ) )
4 df-om
 |-  _om = { y e. On | A. x ( Lim x -> y e. x ) }
5 3 4 elrab2
 |-  ( A e. _om <-> ( A e. On /\ A. x ( Lim x -> A e. x ) ) )