Metamath Proof Explorer


Theorem ddif

Description: Double complement under universal class. Exercise 4.10(s) of Mendelson p. 231. (Contributed by NM, 8-Jan-2002)

Ref Expression
Assertion ddif
|- ( _V \ ( _V \ A ) ) = A

Proof

Step Hyp Ref Expression
1 velcomp
 |-  ( x e. ( _V \ A ) <-> -. x e. A )
2 1 con2bii
 |-  ( x e. A <-> -. x e. ( _V \ A ) )
3 vex
 |-  x e. _V
4 3 biantrur
 |-  ( -. x e. ( _V \ A ) <-> ( x e. _V /\ -. x e. ( _V \ A ) ) )
5 2 4 bitr2i
 |-  ( ( x e. _V /\ -. x e. ( _V \ A ) ) <-> x e. A )
6 5 difeqri
 |-  ( _V \ ( _V \ A ) ) = A