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Theorem disj2 4168
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 3766 . 2 𝐴 ⊆ V
2 reldisj 4163 . 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 1632  Vcvv 3340  cdif 3712  cin 3714  wss 3715  c0 4058
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ral 3055  df-v 3342  df-dif 3718  df-in 3722  df-ss 3729  df-nul 4059
This theorem is referenced by:  ssindif0  4175  intirr  5672  setsres  16103  setscom  16105  f1omvdco3  18069  psgnunilem5  18114  opsrtoslem2  19687  clsconn  21435  cldsubg  22115  uniinn0  29673  imadifxp  29721
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