Metamath Proof Explorer


Theorem alexbii

Description: Biconditional form of aleximi . (Contributed by BJ, 16-Nov-2020)

Ref Expression
Hypothesis alexbii.1
|- ( ph -> ( ps <-> ch ) )
Assertion alexbii
|- ( A. x ph -> ( E. x ps <-> E. x ch ) )

Proof

Step Hyp Ref Expression
1 alexbii.1
 |-  ( ph -> ( ps <-> ch ) )
2 1 biimpd
 |-  ( ph -> ( ps -> ch ) )
3 2 aleximi
 |-  ( A. x ph -> ( E. x ps -> E. x ch ) )
4 1 biimprd
 |-  ( ph -> ( ch -> ps ) )
5 4 aleximi
 |-  ( A. x ph -> ( E. x ch -> E. x ps ) )
6 3 5 impbid
 |-  ( A. x ph -> ( E. x ps <-> E. x ch ) )