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Theorem disj2 3388
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 3089 . 2 𝐴 ⊆ V
2 reldisj 3384 . 2 (𝐴 ⊆ V → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
31, 2ax-mp 5 1 ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))
Colors of variables: wff set class
Syntax hints:  wb 104   = wceq 1316  Vcvv 2660  cdif 3038  cin 3040  wss 3041  c0 3333
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-ral 2398  df-v 2662  df-dif 3043  df-in 3047  df-ss 3054  df-nul 3334
This theorem is referenced by:  ssindif0im  3392  intirr  4895  setsresg  11924  setscom  11926
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