Theorem ddif 1597

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 1496	. . . . 5 ⊢ (x ∈ (V ∖ A) ↔ (x ∈ V ∧ ¬ x ∈ A))
2		visset 1350	. . . . 5 ⊢ x ∈ V
3	1, 2	mpbiran 547	. . . 4 ⊢ (x ∈ (V ∖ A) ↔ ¬ x ∈ A)
4	3	bicon2i 194	. . 3 ⊢ (x ∈ A ↔ ¬ x ∈ (V ∖ A))
5	2	biantrur 544	. . 3 ⊢ (¬ x ∈ (V ∖ A) ↔ (x ∈ V ∧ ¬ x ∈ (V ∖ A)))
6	4, 5	bitr2 152	. 2 ⊢ ((x ∈ V ∧ ¬ x ∈ (V ∖ A)) ↔ x ∈ A)
7	6	difeqri 1589	1 ⊢ (V ∖ (V ∖ A)) = A

Syntax hints:  ¬ wn 1   ∧ wa 196   = wceq 1091   ∈ wcel 1092  Vcvv 1348   ∖ cdif 1484

This theorem is referenced by:  dfun3 1671  dfin3 1672  invdif 1674  ssindif0 1741  difdifdir 1765

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489

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