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 ∈ V
2		eldif 3222	. . . . 5 ⊢ (x ∈ (V ∖ A) ↔ (x ∈ V ∧ ¬ x ∈ A))
3	1, 2	mpbiran 884	. . . 4 ⊢ (x ∈ (V ∖ A) ↔ ¬ x ∈ A)
4	3	con2bii 322	. . 3 ⊢ (x ∈ A ↔ ¬ x ∈ (V ∖ A))
5	1	biantrur 492	. . 3 ⊢ (¬ x ∈ (V ∖ A) ↔ (x ∈ V ∧ ¬ x ∈ (V ∖ A)))
6	4, 5	bitr2i 241	. 2 ⊢ ((x ∈ V ∧ ¬ x ∈ (V ∖ A)) ↔ x ∈ A)
7	6	difeqri 3388	1 ⊢ (V ∖ (V ∖ A)) = A

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 358   = wceq 1642   ∈ wcel 1710  Vcvv 2860   ∖ cdif 3207

This proof depends on 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 proof 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 used by:  dfun3  3494  dfin3  3495  invdif  3497  ssindif0  3605  difdifdir  3638