Metamath Proof Explorer


Theorem r19.2zb

Description: A response to the notion that the condition A =/= (/) can be removed in r19.2z . Interestingly enough, ph does not figure in the left-hand side. (Contributed by Jeff Hankins, 24-Aug-2009)

Ref Expression
Assertion r19.2zb
|- ( A =/= (/) <-> ( A. x e. A ph -> E. x e. A ph ) )

Proof

Step Hyp Ref Expression
1 r19.2z
 |-  ( ( A =/= (/) /\ A. x e. A ph ) -> E. x e. A ph )
2 1 ex
 |-  ( A =/= (/) -> ( A. x e. A ph -> E. x e. A ph ) )
3 noel
 |-  -. x e. (/)
4 3 pm2.21i
 |-  ( x e. (/) -> ph )
5 4 rgen
 |-  A. x e. (/) ph
6 raleq
 |-  ( A = (/) -> ( A. x e. A ph <-> A. x e. (/) ph ) )
7 5 6 mpbiri
 |-  ( A = (/) -> A. x e. A ph )
8 7 necon3bi
 |-  ( -. A. x e. A ph -> A =/= (/) )
9 exsimpl
 |-  ( E. x ( x e. A /\ ph ) -> E. x x e. A )
10 df-rex
 |-  ( E. x e. A ph <-> E. x ( x e. A /\ ph ) )
11 n0
 |-  ( A =/= (/) <-> E. x x e. A )
12 9 10 11 3imtr4i
 |-  ( E. x e. A ph -> A =/= (/) )
13 8 12 ja
 |-  ( ( A. x e. A ph -> E. x e. A ph ) -> A =/= (/) )
14 2 13 impbii
 |-  ( A =/= (/) <-> ( A. x e. A ph -> E. x e. A ph ) )