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Theorem disj3 3266
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 2921 . . . . 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 1356 . 2 (x(x A → ¬ x B) ↔ x(x Ax (AB)))
6 disj1 3264 . 2 ((AB) = ∅ ↔ x(x A → ¬ x B))
7 dfcleq 2031 . 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 1240   = wceq 1242   wcel 1390  cdif 2908  cin 2910  c0 3218
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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-v 2553  df-dif 2914  df-in 2918  df-nul 3219
This theorem is referenced by:  disjel  3268  disj4im  3270  uneqdifeqim  3302  difprsn1  3494  diftpsn3  3496
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