Theorem disj4 4003

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 3999	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eqcom 2628	. 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ (𝐴 ∖ 𝐵) = 𝐴)
3		difss 3721	. . . 4 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
4		dfpss2 3676	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ (𝐴 ∖ 𝐵) = 𝐴))
5	3, 4	mpbiran 952	. . 3 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) = 𝐴)
6	5	con2bii 347	. 2 ⊢ ((𝐴 ∖ 𝐵) = 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)
7	1, 2, 6	3bitri 286	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)

Syntax hints:  ¬ wn 3   ↔ wb 196   = wceq 1480   ∖ cdif 3557   ∩ cin 3559   ⊆ wss 3560   ⊊ wpss 3561  ∅c0 3897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ne 2791  df-ral 2913  df-v 3192  df-dif 3563  df-in 3567  df-ss 3574  df-pss 3576  df-nul 3898

This theorem is referenced by:  marypha1lem  8299  infeq5i  8493  wilthlem2  24729  topdifinffinlem  32866