Metamath Proof Explorer


Theorem elcnvcnvintab

Description: Two ways of saying a set is an element of the converse of the converse of the intersection of a class. (Contributed by RP, 20-Aug-2020)

Ref Expression
Assertion elcnvcnvintab
|- ( A e. `' `' |^| { x | ph } <-> ( A e. ( _V X. _V ) /\ A. x ( ph -> A e. x ) ) )

Proof

Step Hyp Ref Expression
1 cnvcnv
 |-  `' `' |^| { x | ph } = ( |^| { x | ph } i^i ( _V X. _V ) )
2 incom
 |-  ( |^| { x | ph } i^i ( _V X. _V ) ) = ( ( _V X. _V ) i^i |^| { x | ph } )
3 1 2 eqtri
 |-  `' `' |^| { x | ph } = ( ( _V X. _V ) i^i |^| { x | ph } )
4 3 eleq2i
 |-  ( A e. `' `' |^| { x | ph } <-> A e. ( ( _V X. _V ) i^i |^| { x | ph } ) )
5 elinintab
 |-  ( A e. ( ( _V X. _V ) i^i |^| { x | ph } ) <-> ( A e. ( _V X. _V ) /\ A. x ( ph -> A e. x ) ) )
6 4 5 bitri
 |-  ( A e. `' `' |^| { x | ph } <-> ( A e. ( _V X. _V ) /\ A. x ( ph -> A e. x ) ) )