Theorem disj2 4398

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 3946	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4393	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 206   = wceq 1542  Vcvv 3429   ∖ cdif 3886   ∩ cin 3888   ⊆ wss 3889  ∅c0 4273

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-ral 3052  df-v 3431  df-dif 3892  df-in 3896  df-ss 3906  df-nul 4274

This theorem is referenced by:  ssindif0  4404  intirr  6081  setsres  17148  setscom  17150  f1omvdco3  19424  psgnunilem5  19469  opsrtoslem2  22034  clsconn  23395  cldsubg  24076  uniinn0  32620  imadifxp  32671