Theorem ddif 3399

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 2863	. . . . 5 |- x e. _V
2		eldif 3222	. . . . 5 |- (x e. (_V \ A) <-> (x e. _V /\ -. x e. A))
3	1, 2	mpbiran 884	. . . 4 |- (x e. (_V \ A) <-> -. x e. A)
4	3	con2bii 322	. . 3 |- (x e. A <-> -. x e. (_V \ A))
5	1	biantrur 492	. . 3 |- (-. x e. (_V \ A) <-> (x e. _V /\ -. x e. (_V \ A)))
6	4, 5	bitr2i 241	. 2 |- ((x e. _V /\ -. x e. (_V \ A)) <-> x e. A)
7	6	difeqri 3388	1 |- (_V \ (_V \ A)) = A

Syntax hints:  -. wn 3   /\ wa 358   = wceq 1642   e. wcel 1710  _Vcvv 2860   \ cdif 3207

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216

This theorem is referenced by:  dfun3  3494  dfin3  3495  invdif  3497  ssindif0  3605  difdifdir  3638