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Theorem disj3 3595
 Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.)
Assertion
Ref Expression
disj3 ((AB) = A = (A B))

Proof of Theorem disj3
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 pm4.71 611 . . . 4 ((x A → ¬ x B) ↔ (x A ↔ (x A ¬ x B)))
2 eldif 3221 . . . . 5 (x (A B) ↔ (x A ¬ x B))
32bibi2i 304 . . . 4 ((x Ax (A B)) ↔ (x A ↔ (x A ¬ x B)))
41, 3bitr4i 243 . . 3 ((x A → ¬ x B) ↔ (x Ax (A B)))
54albii 1566 . 2 (x(x A → ¬ x B) ↔ x(x Ax (A B)))
6 disj1 3593 . 2 ((AB) = x(x A → ¬ x B))
7 dfcleq 2347 . 2 (A = (A B) ↔ x(x Ax (A B)))
85, 6, 73bitr4i 268 1 ((AB) = A = (A B))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   = wceq 1642   ∈ wcel 1710   ∖ cdif 3206   ∩ cin 3208  ∅c0 3550 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-ral 2619  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-nul 3551 This theorem is referenced by:  disjel  3597  disj4  3599  uneqdifeq  3638  difprsn1  3847  diftpsn3  3849  ssunsn2  3865
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