Theorem disj 3573

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 3226	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2	1	eqeq1i 2246	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅)
3		abeq1 2348	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅ ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅))
4		imnan 701	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
5		noel 3525	. . . . . 6 ⊢ ¬ 𝑥 ∈ ∅
6	5	nbn 711	. . . . 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 1523	. . 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 2533	. 2 ⊢ (∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
11	9, 10	bitr4i 187	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1400   = wceq 1402   ∈ wcel 2209  {cab 2224  ∀wral 2528   ∩ cin 3219  ∅c0 3520

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-v 2823  df-dif 3222  df-in 3226  df-nul 3521

This theorem is used by:  disjr  3574  disj1  3575  disjne  3578  f0rn0  5587  renfdisj  8385  fvinim0ffz  10670  xnn0nnen  10887  fxnn0nninf  10889  fprodsplitdc  12379  exmidunben  13366  dedekindeulemuub  15767  dedekindeulemlu  15771  dedekindicclemuub  15776  dedekindicclemlu  15780  ivthinclemdisj  15790  exmidsbthrlem  17165