Theorem disj4 4286

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 4281	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eqcom 2780	. 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ (𝐴 ∖ 𝐵) = 𝐴)
3		difss 3993	. . . 4 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
4		dfpss2 3947	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ (𝐴 ∖ 𝐵) = 𝐴))
5	3, 4	mpbiran 697	. . 3 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) = 𝐴)
6	5	con2bii 350	. 2 ⊢ ((𝐴 ∖ 𝐵) = 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)
7	1, 2, 6	3bitri 289	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)

Syntax hints:  ¬ wn 3   ↔ wb 198   = wceq 1508   ∖ cdif 3821   ∩ cin 3823   ⊆ wss 3824   ⊊ wpss 3825  ∅c0 4173

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1759  ax-4 1773  ax-5 1870  ax-6 1929  ax-7 1966  ax-8 2053  ax-9 2060  ax-10 2080  ax-11 2094  ax-12 2107  ax-ext 2745

This theorem depends on definitions:  df-bi 199  df-an 388  df-or 835  df-tru 1511  df-ex 1744  df-nf 1748  df-sb 2017  df-clab 2754  df-cleq 2766  df-clel 2841  df-nfc 2913  df-ne 2963  df-ral 3088  df-v 3412  df-dif 3827  df-in 3831  df-ss 3838  df-pss 3840  df-nul 4174

This theorem is referenced by:  marypha1lem  8691  infeq5i  8892  wilthlem2  25364  topdifinffinlem  34103