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Theorem disj3 3420
Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.)
Assertion
Ref Expression
disj3 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))

Proof of Theorem disj3
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 pm4.71 387 . . . 4 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
2 eldif 3085 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
32bibi2i 226 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐴𝐵)) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
41, 3bitr4i 186 . . 3 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐴𝐵)))
54albii 1447 . 2 (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
6 disj1 3418 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
7 dfcleq 2134 . 2 (𝐴 = (𝐴𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
85, 6, 73bitr4i 211 1 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wb 104  wal 1330   = wceq 1332  wcel 1481  cdif 3073  cin 3075  c0 3368
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-v 2691  df-dif 3078  df-in 3082  df-nul 3369
This theorem is referenced by:  disjel  3422  uneqdifeqim  3453  difprsn1  3667  diftpsn3  3669  orddif  4470  phpm  6767
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