Theorem disj2 4407

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 3991	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4402	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 208   = wceq 1533  Vcvv 3495   ∖ cdif 3933   ∩ cin 3935   ⊆ wss 3936  ∅c0 4291

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2156  ax-12 2172  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-v 3497  df-dif 3939  df-in 3943  df-ss 3952  df-nul 4292

This theorem is referenced by:  ssindif0  4413  intirr  5973  setsres  16519  setscom  16521  f1omvdco3  18571  psgnunilem5  18616  opsrtoslem2  20259  clsconn  22032  cldsubg  22713  uniinn0  30296  imadifxp  30345