Metamath Proof Explorer


Theorem nfceqi

Description: Equality theorem for class not-free. (Contributed by Mario Carneiro, 11-Aug-2016) (Proof shortened by Wolf Lammen, 16-Nov-2019) Avoid ax-12 . (Revised by Wolf Lammen, 19-Jun-2023)

Ref Expression
Hypothesis nfceqi.1
|- A = B
Assertion nfceqi
|- ( F/_ x A <-> F/_ x B )

Proof

Step Hyp Ref Expression
1 nfceqi.1
 |-  A = B
2 1 eleq2i
 |-  ( y e. A <-> y e. B )
3 2 nfbii
 |-  ( F/ x y e. A <-> F/ x y e. B )
4 3 albii
 |-  ( A. y F/ x y e. A <-> A. y F/ x y e. B )
5 df-nfc
 |-  ( F/_ x A <-> A. y F/ x y e. A )
6 df-nfc
 |-  ( F/_ x B <-> A. y F/ x y e. B )
7 4 5 6 3bitr4i
 |-  ( F/_ x A <-> F/_ x B )