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Theorem disj2 3357
 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 3061 . 2 𝐴 ⊆ V
2 reldisj 3353 . 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 104   = wceq 1296  Vcvv 2633   ∖ cdif 3010   ∩ cin 3012   ⊆ wss 3013  ∅c0 3302 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077 This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-ral 2375  df-v 2635  df-dif 3015  df-in 3019  df-ss 3026  df-nul 3303 This theorem is referenced by:  ssindif0im  3361  intirr  4851  setsresg  11681  setscom  11683
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