Theorem disjr 3458

Description: Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.)

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem disjr

Step	Hyp	Ref	Expression
1		incom 3314	. . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	eqeq1i 2173	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)
3		disj 3457	. 2 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)
4	2, 3	bitri 183	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)

Syntax hints:  ¬ wn 3   ↔ wb 104   = wceq 1343   ∈ wcel 2136  ∀wral 2444   ∩ cin 3115  ∅c0 3409

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-v 2728  df-dif 3118  df-in 3122  df-nul 3410

This theorem is referenced by: (None)