Theorem disj2 4391

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 3945	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4385	. 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 3432   ∖ cdif 3884   ∩ cin 3886   ⊆ wss 3887  ∅c0 4256

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904  df-nul 4257

This theorem is referenced by:  ssindif0  4397  intirr  6023  setsres  16879  setscom  16881  f1omvdco3  19057  psgnunilem5  19102  opsrtoslem2  21263  clsconn  22581  cldsubg  23262  uniinn0  30890  imadifxp  30940