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Theorem disj2 3275
 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 2965 . 2 𝐴 ⊆ V
2 reldisj 3271 . 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 1243  Vcvv 2557   ∖ cdif 2914   ∩ cin 2916   ⊆ wss 2917  ∅c0 3224 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 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-v 2559  df-dif 2920  df-in 2924  df-ss 2931  df-nul 3225 This theorem is referenced by:  ssindif0im  3281  intirr  4711
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