Theorem disj4 4408

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 4403	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eqcom 2740	. 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ (𝐴 ∖ 𝐵) = 𝐴)
3		difss 4085	. . . 4 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
4		dfpss2 4037	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ (𝐴 ∖ 𝐵) = 𝐴))
5	3, 4	mpbiran 709	. . 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 1541   ∖ cdif 3895   ∩ cin 3897   ⊆ wss 3898   ⊊ wpss 3899  ∅c0 4282

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ne 2930  df-ral 3049  df-v 3439  df-dif 3901  df-in 3905  df-ss 3915  df-pss 3918  df-nul 4283

This theorem is referenced by:  marypha1lem  9324  infeq5i  9533  wilthlem2  27007  topdifinffinlem  37412