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Theorem disj4 3599
 Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)
Assertion
Ref Expression
disj4 ((AB) = ↔ ¬ (A B) ⊊ A)

Proof of Theorem disj4
StepHypRef Expression
1 disj3 3595 . 2 ((AB) = A = (A B))
2 eqcom 2355 . 2 (A = (A B) ↔ (A B) = A)
3 difss 3393 . . . 4 (A B) A
4 dfpss2 3354 . . . 4 ((A B) ⊊ A ↔ ((A B) A ¬ (A B) = A))
53, 4mpbiran 884 . . 3 ((A B) ⊊ A ↔ ¬ (A B) = A)
65con2bii 322 . 2 ((A B) = A ↔ ¬ (A B) ⊊ A)
71, 2, 63bitri 262 1 ((AB) = ↔ ¬ (A B) ⊊ A)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 176   = wceq 1642   ∖ cdif 3206   ∩ cin 3208   ⊆ wss 3257   ⊊ wpss 3258  ∅c0 3550 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-ral 2619  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259  df-pss 3261  df-nul 3551 This theorem is referenced by: (None)
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