ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  disj3 GIF version

Theorem disj3 3512
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 pm4.71 389 . . . 4 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
2 eldif 3174 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
32bibi2i 227 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐴𝐵)) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
41, 3bitr4i 187 . . 3 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐴𝐵)))
54albii 1492 . 2 (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
6 disj1 3510 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
7 dfcleq 2198 . 2 (𝐴 = (𝐴𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
85, 6, 73bitr4i 212 1 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wb 105  wal 1370   = wceq 1372  wcel 2175  cdif 3162  cin 3164  c0 3459
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186
This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-ral 2488  df-v 2773  df-dif 3167  df-in 3171  df-nul 3460
This theorem is referenced by:  disjel  3514  uneqdifeqim  3545  difprsn1  3771  diftpsn3  3773  orddif  4593  phpm  6944
  Copyright terms: Public domain W3C validator