Metamath Proof Explorer


Theorem bj-nfcf

Description: Version of df-nfc with a disjoint variable condition replaced with a nonfreeness hypothesis. (Contributed by BJ, 2-May-2019)

Ref Expression
Hypothesis bj-nfcf.nf
|- F/_ y A
Assertion bj-nfcf
|- ( F/_ x A <-> A. y F/ x y e. A )

Proof

Step Hyp Ref Expression
1 bj-nfcf.nf
 |-  F/_ y A
2 df-nfc
 |-  ( F/_ x A <-> A. z F/ x z e. A )
3 1 nfcri
 |-  F/ y z e. A
4 3 nfnf
 |-  F/ y F/ x z e. A
5 4 sb8v
 |-  ( A. z F/ x z e. A <-> A. y [ y / z ] F/ x z e. A )
6 bj-sbnf
 |-  ( [ y / z ] F/ x z e. A <-> F/ x [ y / z ] z e. A )
7 clelsb3
 |-  ( [ y / z ] z e. A <-> y e. A )
8 7 nfbii
 |-  ( F/ x [ y / z ] z e. A <-> F/ x y e. A )
9 6 8 bitri
 |-  ( [ y / z ] F/ x z e. A <-> F/ x y e. A )
10 9 albii
 |-  ( A. y [ y / z ] F/ x z e. A <-> A. y F/ x y e. A )
11 5 10 bitri
 |-  ( A. z F/ x z e. A <-> A. y F/ x y e. A )
12 2 11 bitri
 |-  ( F/_ x A <-> A. y F/ x y e. A )