Theorem disj4 4399

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 4394	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eqcom 2743	. 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ (𝐴 ∖ 𝐵) = 𝐴)
3		difss 4076	. . . 4 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
4		dfpss2 4028	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ (𝐴 ∖ 𝐵) = 𝐴))
5	3, 4	mpbiran 710	. . 3 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) = 𝐴)
6	5	con2bii 357	. 2 ⊢ ((𝐴 ∖ 𝐵) = 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)
7	1, 2, 6	3bitri 297	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)

Syntax hints:  ¬ wn 3   ↔ wb 206   = wceq 1542   ∖ cdif 3886   ∩ cin 3888   ⊆ wss 3889   ⊊ wpss 3890  ∅c0 4273

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-ne 2933  df-ral 3052  df-v 3431  df-dif 3892  df-in 3896  df-ss 3906  df-pss 3909  df-nul 4274

This theorem is referenced by:  marypha1lem  9346  infeq5i  9557  wilthlem2  27032  topdifinffinlem  37663