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Theorem disj3 3272
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 pm4.71 369 . . . 4 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
2 eldif 2927 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
32bibi2i 216 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐴𝐵)) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
41, 3bitr4i 176 . . 3 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐴𝐵)))
54albii 1359 . 2 (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
6 disj1 3270 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
7 dfcleq 2034 . 2 (𝐴 = (𝐴𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
85, 6, 73bitr4i 201 1 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 97  wb 98  wal 1241   = wceq 1243  wcel 1393  cdif 2914  cin 2916  c0 3224
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-v 2559  df-dif 2920  df-in 2924  df-nul 3225
This theorem is referenced by:  disjel  3274  disj4im  3276  uneqdifeqim  3308  difprsn1  3503  diftpsn3  3505  orddif  4271  phpm  6327
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