Metamath Proof Explorer


Theorem nfbidv

Description: An equality theorem for nonfreeness. See nfbidf for a version without disjoint variable condition but requiring more axioms. (Contributed by Mario Carneiro, 4-Oct-2016) Remove dependency on ax-6 , ax-7 , ax-12 by adapting proof of nfbidf . (Revised by BJ, 25-Sep-2022)

Ref Expression
Hypothesis albidv.1
|- ( ph -> ( ps <-> ch ) )
Assertion nfbidv
|- ( ph -> ( F/ x ps <-> F/ x ch ) )

Proof

Step Hyp Ref Expression
1 albidv.1
 |-  ( ph -> ( ps <-> ch ) )
2 1 exbidv
 |-  ( ph -> ( E. x ps <-> E. x ch ) )
3 1 albidv
 |-  ( ph -> ( A. x ps <-> A. x ch ) )
4 2 3 imbi12d
 |-  ( ph -> ( ( E. x ps -> A. x ps ) <-> ( E. x ch -> A. x ch ) ) )
5 df-nf
 |-  ( F/ x ps <-> ( E. x ps -> A. x ps ) )
6 df-nf
 |-  ( F/ x ch <-> ( E. x ch -> A. x ch ) )
7 4 5 6 3bitr4g
 |-  ( ph -> ( F/ x ps <-> F/ x ch ) )