Metamath Proof Explorer


Theorem bj-cbvaldv

Description: Version of cbvald with a disjoint variable condition, which does not require ax-13 . (Contributed by BJ, 16-Jun-2019) (Proof modification is discouraged.)

Ref Expression
Hypotheses bj-cbvaldv.1
|- F/ y ph
bj-cbvaldv.2
|- ( ph -> F/ y ps )
bj-cbvaldv.3
|- ( ph -> ( x = y -> ( ps <-> ch ) ) )
Assertion bj-cbvaldv
|- ( ph -> ( A. x ps <-> A. y ch ) )

Proof

Step Hyp Ref Expression
1 bj-cbvaldv.1
 |-  F/ y ph
2 bj-cbvaldv.2
 |-  ( ph -> F/ y ps )
3 bj-cbvaldv.3
 |-  ( ph -> ( x = y -> ( ps <-> ch ) ) )
4 nfv
 |-  F/ x ph
5 nfv
 |-  F/ x ch
6 5 a1i
 |-  ( ph -> F/ x ch )
7 4 1 2 6 3 bj-cbv2v
 |-  ( ph -> ( A. x ps <-> A. y ch ) )