Metamath Proof Explorer


Theorem nfceqdf

Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 14-Oct-2016) Avoid ax-8 and df-clel . (Revised by WL and SN, 23-Aug-2024)

Ref Expression
Hypotheses nfceqdf.1
|- F/ x ph
nfceqdf.2
|- ( ph -> A = B )
Assertion nfceqdf
|- ( ph -> ( F/_ x A <-> F/_ x B ) )

Proof

Step Hyp Ref Expression
1 nfceqdf.1
 |-  F/ x ph
2 nfceqdf.2
 |-  ( ph -> A = B )
3 eleq2w2
 |-  ( A = B -> ( y e. A <-> y e. B ) )
4 2 3 syl
 |-  ( ph -> ( y e. A <-> y e. B ) )
5 1 4 nfbidf
 |-  ( ph -> ( F/ x y e. A <-> F/ x y e. B ) )
6 5 albidv
 |-  ( ph -> ( A. y F/ x y e. A <-> A. y F/ x y e. B ) )
7 df-nfc
 |-  ( F/_ x A <-> A. y F/ x y e. A )
8 df-nfc
 |-  ( F/_ x B <-> A. y F/ x y e. B )
9 6 7 8 3bitr4g
 |-  ( ph -> ( F/_ x A <-> F/_ x B ) )