Theorem ddif 3742

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

Assertion

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

Proof of Theorem ddif

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3203	. . . . 5 |- x e. _V
2		eldif 3584	. . . . 5 |- ( x e. ( _V \ A ) <-> ( x e. _V /\ -. x e. A ) )
3	1, 2	mpbiran 953	. . . 4 |- ( x e. ( _V \ A ) <-> -. x e. A )
4	3	con2bii 347	. . 3 |- ( x e. A <-> -. x e. ( _V \ A ) )
5	1	biantrur 527	. . 3 |- ( -. x e. ( _V \ A ) <-> ( x e. _V /\ -. x e. ( _V \ A ) ) )
6	4, 5	bitr2i 265	. 2 |- ( ( x e. _V /\ -. x e. ( _V \ A ) ) <-> x e. A )
7	6	difeqri 3730	1 |- ( _V \ ( _V \ A ) ) = A

Syntax hints:   -. wn 3   /\ wa 384   = wceq 1483   e. wcel 1990   _Vcvv 3200   \ cdif 3571

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577

This theorem is referenced by:  complss  3751  dfun3  3865  dfin3  3866  invdif  3868  ssindif0  4031  difdifdir  4056