Theorem disj2 4411

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 3955	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4406	. 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 3451   ∖ cdif 3896   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279

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 2147  ax-9 2155  ax-ext 2733

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 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-v 3453  df-dif 3902  df-in 3906  df-ss 3916  df-nul 4280

This theorem is used by:  ssindif0  4417  intirr  6110  setsres  17336  setscom  17338  f1omvdco3  19643  psgnunilem5  19688  opsrtoslem2  22345  clsconn  23728  cldsubg  24410  uniinn0  33129  imadifxp  33177