Metamath Proof Explorer


Theorem spcimegf

Description: Existential specialization, using implicit substitution. (Contributed by Mario Carneiro, 4-Jan-2017)

Ref Expression
Hypotheses spcimgf.1
|- F/_ x A
spcimgf.2
|- F/ x ps
spcimegf.3
|- ( x = A -> ( ps -> ph ) )
Assertion spcimegf
|- ( A e. V -> ( ps -> E. x ph ) )

Proof

Step Hyp Ref Expression
1 spcimgf.1
 |-  F/_ x A
2 spcimgf.2
 |-  F/ x ps
3 spcimegf.3
 |-  ( x = A -> ( ps -> ph ) )
4 2 nfn
 |-  F/ x -. ps
5 3 con3d
 |-  ( x = A -> ( -. ph -> -. ps ) )
6 1 4 5 spcimgf
 |-  ( A e. V -> ( A. x -. ph -> -. ps ) )
7 6 con2d
 |-  ( A e. V -> ( ps -> -. A. x -. ph ) )
8 df-ex
 |-  ( E. x ph <-> -. A. x -. ph )
9 7 8 syl6ibr
 |-  ( A e. V -> ( ps -> E. x ph ) )