Theorem disj5 3891

Description: Two ways of saying that two classes are disjoint. (Contributed by SF, 5-Feb-2015.)

Assertion

Ref	Expression
disj5	⊢ ((A ∩ B) = ∅ ↔ A ⊆ ∼ B)

Proof of Theorem disj5

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 2863	. . . . 5 ⊢ x ∈ V
2	1	elcompl 3226	. . . 4 ⊢ (x ∈ ∼ B ↔ ¬ x ∈ B)
3	2	ralbii 2639	. . 3 ⊢ (∀x ∈ A x ∈ ∼ B ↔ ∀x ∈ A ¬ x ∈ B)
4		df-ral 2620	. . 3 ⊢ (∀x ∈ A x ∈ ∼ B ↔ ∀x(x ∈ A → x ∈ ∼ B))
5	3, 4	bitr3i 242	. 2 ⊢ (∀x ∈ A ¬ x ∈ B ↔ ∀x(x ∈ A → x ∈ ∼ B))
6		disj 3592	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)
7		dfss2 3263	. 2 ⊢ (A ⊆ ∼ B ↔ ∀x(x ∈ A → x ∈ ∼ B))
8	5, 6, 7	3bitr4i 268	1 ⊢ ((A ∩ B) = ∅ ↔ A ⊆ ∼ B)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176  ∀wal 1540   = wceq 1642   ∈ wcel 1710  ∀wral 2615   ∼ ccompl 3206   ∩ 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:  intirr  5030