Theorem ddif 2169

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 2057	. . . . 5 |- (x e. (V \ A) <-> (x e. V /\ -. x e. A))
2		visset 1813	. . . . 5 |- x e. V
3	1, 2	mpbiran 728	. . . 4 |- (x e. (V \ A) <-> -. x e. A)
4	3	con2bii 221	. . 3 |- (x e. A <-> -. x e. (V \ A))
5	2	biantrur 725	. . 3 |- (-. x e. (V \ A) <-> (x e. V /\ -. x e. (V \ A)))
6	4, 5	bitr2 174	. 2 |- ((x e. V /\ -. x e. (V \ A)) <-> x e. A)
7	6	difeqri 2160	1 |- (V \ (V \ A)) = A

Syntax hints:  -. wn 2   /\ wa 223   = wceq 956   e. wcel 958  Vcvv 1811   \ cdif 2044

This theorem is referenced by:  dfun3 2246  dfin3 2247  invdif 2249  ssindif0 2322  difdifdir 2346

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049

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