Theorem disj2 4414

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
disj2	⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Proof of Theorem disj2

Step	Hyp	Ref	Expression
1		ssv 3937	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4409	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570  Vcvv 3454   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282

This theorem is used by:  ssindif0  4420  intirr  6124  setsres  17385  setscom  17387  f1omvdco3  19693  psgnunilem5  19738  opsrtoslem2  22395  clsconn  23778  cldsubg  24460  uniinn0  33219  imadifxp  33267