Metamath Proof Explorer


Theorem abid2f

Description: A simplification of class abstraction. Theorem 5.2 of Quine p. 35. (Contributed by NM, 5-Sep-2011) (Revised by Mario Carneiro, 7-Oct-2016) (Proof shortened by Wolf Lammen, 17-Nov-2019)

Ref Expression
Hypothesis abid2f.1
|- F/_ x A
Assertion abid2f
|- { x | x e. A } = A

Proof

Step Hyp Ref Expression
1 abid2f.1
 |-  F/_ x A
2 nfab1
 |-  F/_ x { x | x e. A }
3 2 1 cleqf
 |-  ( { x | x e. A } = A <-> A. x ( x e. { x | x e. A } <-> x e. A ) )
4 abid
 |-  ( x e. { x | x e. A } <-> x e. A )
5 3 4 mpgbir
 |-  { x | x e. A } = A