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Theorem disj4 4169
Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)
Assertion
Ref Expression
disj4 ((𝐴𝐵) = ∅ ↔ ¬ (𝐴𝐵) ⊊ 𝐴)

Proof of Theorem disj4
StepHypRef Expression
1 disj3 4164 . 2 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
2 eqcom 2767 . 2 (𝐴 = (𝐴𝐵) ↔ (𝐴𝐵) = 𝐴)
3 difss 3880 . . . 4 (𝐴𝐵) ⊆ 𝐴
4 dfpss2 3834 . . . 4 ((𝐴𝐵) ⊊ 𝐴 ↔ ((𝐴𝐵) ⊆ 𝐴 ∧ ¬ (𝐴𝐵) = 𝐴))
53, 4mpbiran 991 . . 3 ((𝐴𝐵) ⊊ 𝐴 ↔ ¬ (𝐴𝐵) = 𝐴)
65con2bii 346 . 2 ((𝐴𝐵) = 𝐴 ↔ ¬ (𝐴𝐵) ⊊ 𝐴)
71, 2, 63bitri 286 1 ((𝐴𝐵) = ∅ ↔ ¬ (𝐴𝐵) ⊊ 𝐴)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196   = wceq 1632  cdif 3712  cin 3714  wss 3715  wpss 3716  c0 4058
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ne 2933  df-ral 3055  df-v 3342  df-dif 3718  df-in 3722  df-ss 3729  df-pss 3731  df-nul 4059
This theorem is referenced by:  marypha1lem  8504  infeq5i  8706  wilthlem2  24994  topdifinffinlem  33506
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