Theorem disj 3262

Description: Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 17-Feb-2004.)

Assertion

Ref	Expression
disj	⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)

Distinct variable groups:   x,A   x,B

Proof of Theorem disj

Step	Hyp	Ref	Expression
1		df-in 2918	. . . 4 ⊢ (A ∩ B) = {x ∣ (x ∈ A ∧ x ∈ B)}
2	1	eqeq1i 2044	. . 3 ⊢ ((A ∩ B) = ∅ ↔ {x ∣ (x ∈ A ∧ x ∈ B)} = ∅)
3		abeq1 2144	. . 3 ⊢ ({x ∣ (x ∈ A ∧ x ∈ B)} = ∅ ↔ ∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅))
4		imnan 623	. . . . 5 ⊢ ((x ∈ A → ¬ x ∈ B) ↔ ¬ (x ∈ A ∧ x ∈ B))
5		noel 3222	. . . . . 6 ⊢ ¬ x ∈ ∅
6	5	nbn 614	. . . . 5 ⊢ (¬ (x ∈ A ∧ x ∈ B) ↔ ((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅))
7	4, 6	bitr2i 174	. . . 4 ⊢ (((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ (x ∈ A → ¬ x ∈ B))
8	7	albii 1356	. . 3 ⊢ (∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ ∀x(x ∈ A → ¬ x ∈ B))
9	2, 3, 8	3bitri 195	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))
10		df-ral 2305	. 2 ⊢ (∀x ∈ A ¬ x ∈ B ↔ ∀x(x ∈ A → ¬ x ∈ B))
11	9, 10	bitr4i 176	1 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240   = wceq 1242   ∈ wcel 1390  {cab 2023  ∀wral 2300   ∩ cin 2910  ∅c0 3218

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-v 2553  df-dif 2914  df-in 2918  df-nul 3219

This theorem is referenced by:  disjr  3263  disj1  3264  disjne  3267  renfdisj  6856