Theorem disj2 4415

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 3963	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 4410	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 209   = wceq 1563  Vcvv 3457   ∖ cdif 3904   ∩ cin 3906   ⊆ wss 3907  ∅c0 4288

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3080  df-v 3459  df-dif 3910  df-in 3914  df-ss 3924  df-nul 4289

This theorem is referenced by:  ssindif0  4421  intirr  6109  setsres  17228  setscom  17230  f1omvdco3  19510  psgnunilem5  19555  opsrtoslem2  22167  clsconn  23548  cldsubg  24229  uniinn0  32807  imadifxp  32856