Theorem disj2 3599

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 3292	. 2 ⊢ A ⊆ V
2		reldisj 3595	. 2 ⊢ (A ⊆ V → ((A ∩ B) = ∅ ↔ A ⊆ (V ∖ B)))
3	1, 2	ax-mp 5	1 ⊢ ((A ∩ B) = ∅ ↔ A ⊆ (V ∖ B))

Syntax hints:   ↔ wb 176   = wceq 1642  Vcvv 2860   ∖ cdif 3207   ∩ cin 3209   ⊆ wss 3258  ∅c0 3551

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by:  ssindif0  3605