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Theorem disj4 4003
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 3999 . 2 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
2 eqcom 2628 . 2 (𝐴 = (𝐴𝐵) ↔ (𝐴𝐵) = 𝐴)
3 difss 3721 . . . 4 (𝐴𝐵) ⊆ 𝐴
4 dfpss2 3676 . . . 4 ((𝐴𝐵) ⊊ 𝐴 ↔ ((𝐴𝐵) ⊆ 𝐴 ∧ ¬ (𝐴𝐵) = 𝐴))
53, 4mpbiran 952 . . 3 ((𝐴𝐵) ⊊ 𝐴 ↔ ¬ (𝐴𝐵) = 𝐴)
65con2bii 347 . 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 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
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