Metamath Proof Explorer


Theorem eu6im

Description: One direction of eu6 needs fewer axioms. (Contributed by Wolf Lammen, 2-Mar-2023)

Ref Expression
Assertion eu6im
|- ( E. y A. x ( ph <-> x = y ) -> E! x ph )

Proof

Step Hyp Ref Expression
1 exsbim
 |-  ( E. y A. x ( x = y -> ph ) -> E. x ph )
2 1 anim1i
 |-  ( ( E. y A. x ( x = y -> ph ) /\ E. z A. x ( ph -> x = z ) ) -> ( E. x ph /\ E. z A. x ( ph -> x = z ) ) )
3 eu6lem
 |-  ( E. y A. x ( ph <-> x = y ) <-> ( E. y A. x ( x = y -> ph ) /\ E. z A. x ( ph -> x = z ) ) )
4 eu3v
 |-  ( E! x ph <-> ( E. x ph /\ E. z A. x ( ph -> x = z ) ) )
5 2 3 4 3imtr4i
 |-  ( E. y A. x ( ph <-> x = y ) -> E! x ph )