Theorem disj2 4410

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 3932	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4405	. 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 1569  Vcvv 3450   ⊆ wss 3895   ∖ cdif 4045   ∩ cin 4089  ∅c0 4278

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-nul 4279

This theorem is used by:  ssindif0  4416  intirr  6111  setsres  17361  setscom  17363  f1omvdco3  19668  psgnunilem5  19713  opsrtoslem2  22370  clsconn  23753  cldsubg  24435  uniinn0  33194  imadifxp  33242