Metamath Proof Explorer


Theorem iinxsng

Description: A singleton index picks out an instance of an indexed intersection's argument. (Contributed by NM, 15-Jan-2012) (Proof shortened by Mario Carneiro, 17-Nov-2016)

Ref Expression
Hypothesis iinxsng.1
|- ( x = A -> B = C )
Assertion iinxsng
|- ( A e. V -> |^|_ x e. { A } B = C )

Proof

Step Hyp Ref Expression
1 iinxsng.1
 |-  ( x = A -> B = C )
2 df-iin
 |-  |^|_ x e. { A } B = { y | A. x e. { A } y e. B }
3 1 eleq2d
 |-  ( x = A -> ( y e. B <-> y e. C ) )
4 3 ralsng
 |-  ( A e. V -> ( A. x e. { A } y e. B <-> y e. C ) )
5 4 abbi1dv
 |-  ( A e. V -> { y | A. x e. { A } y e. B } = C )
6 2 5 eqtrid
 |-  ( A e. V -> |^|_ x e. { A } B = C )