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Theorem disj2 4015
 Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)
Assertion
Ref Expression
disj2 ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Proof of Theorem disj2
StepHypRef Expression
1 ssv 3617 . 2 𝐴 ⊆ V
2 reldisj 4011 . 2 (𝐴 ⊆ V → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
31, 2ax-mp 5 1 ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 196   = wceq 1481  Vcvv 3195   ∖ cdif 3564   ∩ cin 3566   ⊆ wss 3567  ∅c0 3907 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1484  df-ex 1703  df-nf 1708  df-sb 1879  df-clab 2607  df-cleq 2613  df-clel 2616  df-nfc 2751  df-ral 2914  df-v 3197  df-dif 3570  df-in 3574  df-ss 3581  df-nul 3908 This theorem is referenced by:  ssindif0  4022  intirr  5502  setsres  15882  setscom  15884  f1omvdco3  17850  psgnunilem5  17895  opsrtoslem2  19466  clsconn  21214  cldsubg  21895  uniinn0  29338  imadifxp  29386
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