Metamath Proof Explorer


Theorem abeq1

Description: Equality of a class variable and a class abstraction. Commuted form of abeq2 . (Contributed by NM, 20-Aug-1993)

Ref Expression
Assertion abeq1
|- ( { x | ph } = A <-> A. x ( ph <-> x e. A ) )

Proof

Step Hyp Ref Expression
1 abeq2
 |-  ( A = { x | ph } <-> A. x ( x e. A <-> ph ) )
2 eqcom
 |-  ( { x | ph } = A <-> A = { x | ph } )
3 bicom
 |-  ( ( ph <-> x e. A ) <-> ( x e. A <-> ph ) )
4 3 albii
 |-  ( A. x ( ph <-> x e. A ) <-> A. x ( x e. A <-> ph ) )
5 1 2 4 3bitr4i
 |-  ( { x | ph } = A <-> A. x ( ph <-> x e. A ) )