Metamath Proof Explorer


Theorem ax9v1

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

Ref Expression
Assertion ax9v1
|- ( x = y -> ( z e. x -> z e. y ) )

Proof

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