Theorem disj2 4388

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 3941	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4382	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 205   = wceq 1539  Vcvv 3422   ∖ cdif 3880   ∩ cin 3882   ⊆ wss 3883  ∅c0 4253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900  df-nul 4254

This theorem is referenced by:  ssindif0  4394  intirr  6012  setsres  16807  setscom  16809  f1omvdco3  18972  psgnunilem5  19017  opsrtoslem2  21173  clsconn  22489  cldsubg  23170  uniinn0  30791  imadifxp  30841