Theorem disjr 3593

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 3449	. . 3 ⊢ (A ∩ B) = (B ∩ A)
2	1	eqeq1i 2360	. 2 ⊢ ((A ∩ B) = ∅ ↔ (B ∩ A) = ∅)
3		disj 3592	. 2 ⊢ ((B ∩ A) = ∅ ↔ ∀x ∈ B ¬ x ∈ A)
4	2, 3	bitri 240	1 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ B ¬ x ∈ A)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 176   = wceq 1642   ∈ wcel 1710  ∀wral 2615   ∩ cin 3209  ∅c0 3551

This proof depends on 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 proof 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-nul 3552

This theorem is used by:  nnadjoinpw  4522  sfinltfin  4536