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Theorem disj2 3269
 Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)
Assertion
Ref Expression
disj2 ((AB) = ∅ ↔ A ⊆ (V ∖ B))

Proof of Theorem disj2
StepHypRef Expression
1 ssv 2959 . 2 A ⊆ V
2 reldisj 3265 . 2 (A ⊆ V → ((AB) = ∅ ↔ A ⊆ (V ∖ B)))
31, 2ax-mp 7 1 ((AB) = ∅ ↔ A ⊆ (V ∖ B))
 Colors of variables: wff set class Syntax hints:   ↔ wb 98   = wceq 1242  Vcvv 2551   ∖ cdif 2908   ∩ cin 2910   ⊆ wss 2911  ∅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-ss 2925  df-nul 3219 This theorem is referenced by:  ssindif0im  3275  intirr  4654
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