Metamath Proof Explorer


Theorem vjust

Description: Justification theorem for df-v . (Contributed by Rodolfo Medina, 27-Apr-2010)

Ref Expression
Assertion vjust
|- { x | x = x } = { y | y = y }

Proof

Step Hyp Ref Expression
1 equid
 |-  x = x
2 1 vexw
 |-  z e. { x | x = x }
3 equid
 |-  y = y
4 3 vexw
 |-  z e. { y | y = y }
5 2 4 2th
 |-  ( z e. { x | x = x } <-> z e. { y | y = y } )
6 5 eqriv
 |-  { x | x = x } = { y | y = y }