Metamath Proof Explorer


Theorem ax9v2

Description: Second of two weakened versions of ax9v , with an extra disjoint variable condition on y , z see comments there. (Contributed by BJ, 7-Dec-2020)

Ref Expression
Assertion ax9v2
|- ( x = y -> ( z e. x -> z e. y ) )

Proof

Step Hyp Ref Expression
1 ax9v
 |-  ( x = y -> ( z e. x -> z e. y ) )