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Theorem disj3 3303
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 pm4.71 381 . . . 4 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
2 eldif 2983 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
32bibi2i 225 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐴𝐵)) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
41, 3bitr4i 185 . . 3 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐴𝐵)))
54albii 1400 . 2 (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
6 disj1 3301 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
7 dfcleq 2076 . 2 (𝐴 = (𝐴𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
85, 6, 73bitr4i 210 1 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wb 103  wal 1283   = wceq 1285  wcel 1434  cdif 2971  cin 2973  c0 3258
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-ral 2354  df-v 2604  df-dif 2976  df-in 2980  df-nul 3259
This theorem is referenced by:  disjel  3305  uneqdifeqim  3335  difprsn1  3533  diftpsn3  3535  orddif  4298  phpm  6400
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