Metamath Proof Explorer

Theorem bj-hbaeb

Description: Biconditional version of hbae . (Contributed by BJ, 6-Oct-2018) (Proof modification is discouraged.)

Ref Expression
Assertion bj-hbaeb 
|- ( A. x x = y <-> A. z A. x x = y )

Proof

Step Hyp Ref Expression
1 bj-hbaeb2 
 |-  ( A. x x = y <-> A. x A. z x = y )
2 alcom 
 |-  ( A. x A. z x = y <-> A. z A. x x = y )
3 1 2 bitri 
 |-  ( A. x x = y <-> A. z A. x x = y )