Theorem disj 3500

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 3163	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2	1	eqeq1i 2204	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅)
3		abeq1 2306	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅ ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅))
4		imnan 691	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
5		noel 3455	. . . . . 6 ⊢ ¬ 𝑥 ∈ ∅
6	5	nbn 700	. . . . 5 ⊢ (¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅))
7	4, 6	bitr2i 185	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅) ↔ (𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
8	7	albii 1484	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅) ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
9	2, 3, 8	3bitri 206	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
10		df-ral 2480	. 2 ⊢ (∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
11	9, 10	bitr4i 187	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1362   = wceq 1364   ∈ wcel 2167  {cab 2182  ∀wral 2475   ∩ cin 3156  ∅c0 3451

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-v 2765  df-dif 3159  df-in 3163  df-nul 3452

This theorem is referenced by:  disjr  3501  disj1  3502  disjne  3505  f0rn0  5455  renfdisj  8105  fvinim0ffz  10336  xnn0nnen  10548  fxnn0nninf  10550  fprodsplitdc  11780  exmidunben  12670  dedekindeulemuub  14961  dedekindeulemlu  14965  dedekindicclemuub  14970  dedekindicclemlu  14974  ivthinclemdisj  14984  exmidsbthrlem  15779