Theorem disj4 4415

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 4410	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eqcom 2770	. 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ (𝐴 ∖ 𝐵) = 𝐴)
3		difss 4075	. . . 4 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
4		dfpss2 4018	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ (𝐴 ∖ 𝐵) = 𝐴))
5	3, 4	mpbiran 722	. . 3 ⊢ ((𝐴 ∖ 𝐵) ⊊ 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) = 𝐴)
6	5	con2bii 360	. 2 ⊢ ((𝐴 ∖ 𝐵) = 𝐴 ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)
7	1, 2, 6	3bitri 300	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   = wceq 1570   ⊆ wss 3900   ⊊ wpss 4016   ∖ cdif 4048   ∩ cin 4092  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-v 3456  df-ss 3901  df-pss 4017  df-dif 4050  df-in 4094  df-nul 4282

This theorem is used by:  marypha1lem  9455  infeq5i  9667  wilthlem2  27429  topdifinffinlem  38475