Theorem ddif 2737

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231.

Assertion

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

Proof of Theorem ddif

Step	Hyp	Ref	Expression
1		eldif 2609	. . . . 5 |- (x e. (_V \ A) <-> (x e. _V /\ -. x e. A))
2		visset 2295	. . . . 5 |- x e. _V
3	1, 2	mpbiran 798	. . . 4 |- (x e. (_V \ A) <-> -. x e. A)
4	3	con2bii 238	. . 3 |- (x e. A <-> -. x e. (_V \ A))
5	2	biantrur 794	. . 3 |- (-. x e. (_V \ A) <-> (x e. _V /\ -. x e. (_V \ A)))
6	4, 5	bitr2i 191	. 2 |- ((x e. _V /\ -. x e. (_V \ A)) <-> x e. A)
7	6	difeqri 2727	1 |- (_V \ (_V \ A)) = A

Syntax hints:  -. wn 2   /\ wa 240   = wceq 1298   e. wcel 1300  _Vcvv 2292   \ cdif 2590

This theorem is referenced by:  dfun3 2833  dfin3 2834  invdif 2836  ssindif0 2927  difdifdir 2957

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-dif 2597

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