Metamath Proof Explorer


Theorem ax9

Description: Proof of ax-9 from ax9v1 and ax9v2 , proving sufficiency of the conjunction of the latter two weakened versions of ax9v , which is itself a weakened version of ax-9 . (Contributed by BJ, 7-Dec-2020) (Proof shortened by Wolf Lammen, 11-Apr-2021)

Ref Expression
Assertion ax9
|- ( x = y -> ( z e. x -> z e. y ) )

Proof

Step Hyp Ref Expression
1 equvinv
 |-  ( x = y <-> E. t ( t = x /\ t = y ) )
2 ax9v2
 |-  ( x = t -> ( z e. x -> z e. t ) )
3 2 equcoms
 |-  ( t = x -> ( z e. x -> z e. t ) )
4 ax9v1
 |-  ( t = y -> ( z e. t -> z e. y ) )
5 3 4 sylan9
 |-  ( ( t = x /\ t = y ) -> ( z e. x -> z e. y ) )
6 5 exlimiv
 |-  ( E. t ( t = x /\ t = y ) -> ( z e. x -> z e. y ) )
7 1 6 sylbi
 |-  ( x = y -> ( z e. x -> z e. y ) )