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 elequ1
 |-  ( y = w -> ( y e. x <-> w e. x ) )
2 1 imbi2d
 |-  ( y = w -> ( ( Lim x -> y e. x ) <-> ( Lim x -> w e. x ) ) )
3 2 albidv
 |-  ( y = w -> ( A. x ( Lim x -> y e. x ) <-> A. x ( Lim x -> w e. x ) ) )
4 eleq1
 |-  ( w = A -> ( w e. x <-> A e. x ) )
5 4 imbi2d
 |-  ( w = A -> ( ( Lim x -> w e. x ) <-> ( Lim x -> A e. x ) ) )
6 5 albidv
 |-  ( w = A -> ( A. x ( Lim x -> w e. x ) <-> A. x ( Lim x -> A e. x ) ) )
7 df-om
 |-  _om = { y e. On | A. x ( Lim x -> y e. x ) }
8 3 6 7 elrab2w
 |-  ( A e. _om <-> ( A e. On /\ A. x ( Lim x -> A e. x ) ) )