Metamath Proof Explorer


Theorem abeq2f

Description: Equality of a class variable and a class abstraction. In this version, the fact that x is a non-free variable in A is explicitly stated as a hypothesis. (Contributed by Thierry Arnoux, 11-May-2017) Avoid ax-13 . (Revised by Wolf Lammen, 13-May-2023)

Ref Expression
Hypothesis abeq2f.0
|- F/_ x A
Assertion abeq2f
|- ( A = { x | ph } <-> A. x ( x e. A <-> ph ) )

Proof

Step Hyp Ref Expression
1 abeq2f.0
 |-  F/_ x A
2 nfab1
 |-  F/_ x { x | ph }
3 1 2 cleqf
 |-  ( A = { x | ph } <-> A. x ( x e. A <-> x e. { x | ph } ) )
4 abid
 |-  ( x e. { x | ph } <-> ph )
5 4 bibi2i
 |-  ( ( x e. A <-> x e. { x | ph } ) <-> ( x e. A <-> ph ) )
6 5 albii
 |-  ( A. x ( x e. A <-> x e. { x | ph } ) <-> A. x ( x e. A <-> ph ) )
7 3 6 bitri
 |-  ( A = { x | ph } <-> A. x ( x e. A <-> ph ) )