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Theorem disj5 3890
Description: Two ways of saying that two classes are disjoint. (Contributed by SF, 5-Feb-2015.)
Assertion
Ref Expression
disj5 ((AB) = A B)

Proof of Theorem disj5
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 vex 2862 . . . . 5 x V
21elcompl 3225 . . . 4 (x B ↔ ¬ x B)
32ralbii 2638 . . 3 (x A x Bx A ¬ x B)
4 df-ral 2619 . . 3 (x A x Bx(x Ax B))
53, 4bitr3i 242 . 2 (x A ¬ x Bx(x Ax B))
6 disj 3591 . 2 ((AB) = x A ¬ x B)
7 dfss2 3262 . 2 (A Bx(x Ax B))
85, 6, 73bitr4i 268 1 ((AB) = A B)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 176  wal 1540   = wceq 1642   wcel 1710  wral 2614  ccompl 3205  cin 3208   wss 3257  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-ss 3259  df-nul 3551
This theorem is referenced by:  intirr  5029
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