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

Theorem disj3 3247
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 ((x A → ¬ x B) ↔ (x A ↔ (x A ¬ x B)))
2 eldif 2902 . . . . 5 (x (AB) ↔ (x A ¬ x B))
32bibi2i 216 . . . 4 ((x Ax (AB)) ↔ (x A ↔ (x A ¬ x B)))
41, 3bitr4i 176 . . 3 ((x A → ¬ x B) ↔ (x Ax (AB)))
54albii 1339 . 2 (x(x A → ¬ x B) ↔ x(x Ax (AB)))
6 disj1 3245 . 2 ((AB) = ∅ ↔ x(x A → ¬ x B))
7 dfcleq 2016 . 2 (A = (AB) ↔ x(x Ax (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 1226   = wceq 1228   wcel 1374  cdif 2889  cin 2891  c0 3199
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 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-ral 2287  df-v 2535  df-dif 2895  df-in 2899  df-nul 3200
This theorem is referenced by:  disjel  3249  disj4im  3251  uneqdifeqim  3283  difprsn1  3475  diftpsn3  3477
  Copyright terms: Public domain W3C validator