Theorem disj4 3600

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)

Assertion

Ref	Expression
disj4	⊢ ((A ∩ B) = ∅ ↔ ¬ (A ∖ B) ⊊ A)

Proof of Theorem disj4

Step	Hyp	Ref	Expression
1		disj3 3596	. 2 ⊢ ((A ∩ B) = ∅ ↔ A = (A ∖ B))
2		eqcom 2355	. 2 ⊢ (A = (A ∖ B) ↔ (A ∖ B) = A)
3		difss 3394	. . . 4 ⊢ (A ∖ B) ⊆ A
4		dfpss2 3355	. . . 4 ⊢ ((A ∖ B) ⊊ A ↔ ((A ∖ B) ⊆ A ∧ ¬ (A ∖ B) = A))
5	3, 4	mpbiran 884	. . 3 ⊢ ((A ∖ B) ⊊ A ↔ ¬ (A ∖ B) = A)
6	5	con2bii 322	. 2 ⊢ ((A ∖ B) = A ↔ ¬ (A ∖ B) ⊊ A)
7	1, 2, 6	3bitri 262	1 ⊢ ((A ∩ B) = ∅ ↔ ¬ (A ∖ B) ⊊ A)

Syntax hints:  ¬ wn 3   ↔ wb 176   = wceq 1642   ∖ cdif 3207   ∩ cin 3209   ⊆ wss 3258   ⊊ wpss 3259  ∅c0 3551

This theorem was proved from 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 theorem 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-ss 3260  df-pss 3262  df-nul 3552

This theorem is referenced by: (None)