Theorem disj2 4421

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 3964	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4416	. 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 3458   ∖ cdif 3905   ∩ cin 3907   ⊆ wss 3908  ∅c0 4289

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-v 3460  df-dif 3911  df-in 3915  df-ss 3925  df-nul 4290

This theorem is used by:  ssindif0  4427  intirr  6123  setsres  17263  setscom  17265  f1omvdco3  19550  psgnunilem5  19595  opsrtoslem2  22244  clsconn  23624  cldsubg  24305  uniinn0  32934  imadifxp  32983