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Metamath Proof Explorer

Theorem abrexss

Description: A necessary condition for an image set to be a subset. (Contributed by Thierry Arnoux, 6-Feb-2017)

Ref Expression
Hypothesis abrexss.1 
|- F/_ x C
Assertion abrexss 
|- ( A. x e. A B e. C -> { y | E. x e. A y = B } C_ C )

Proof

Step Hyp Ref Expression
1 abrexss.1 
 |-  F/_ x C
2 nfra1 
 |-  F/ x A. x e. A B e. C
3 1 nfcri 
 |-  F/ x z e. C
4 eleq1 
 |-  ( z = B -> ( z e. C <-> B e. C ) )
5 vex 
 |-  z e. _V
6 5 a1i 
 |-  ( A. x e. A B e. C -> z e. _V )
7 rspa 
 |-  ( ( A. x e. A B e. C /\ x e. A ) -> B e. C )
8 2 3 4 6 7 elabreximd 
 |-  ( ( A. x e. A B e. C /\ z e. { y | E. x e. A y = B } ) -> z e. C )
9 8 ex 
 |-  ( A. x e. A B e. C -> ( z e. { y | E. x e. A y = B } -> z e. C ) )
10 9 ssrdv 
 |-  ( A. x e. A B e. C -> { y | E. x e. A y = B } C_ C )