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Theorem disj3 3410
Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.)
Assertion
Ref Expression
disj3 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))

Proof of Theorem disj3
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 pm4.71 386 . . . 4 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
2 eldif 3075 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
32bibi2i 226 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐴𝐵)) ↔ (𝑥𝐴 ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵)))
41, 3bitr4i 186 . . 3 ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐴𝐵)))
54albii 1446 . 2 (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
6 disj1 3408 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
7 dfcleq 2131 . 2 (𝐴 = (𝐴𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐴𝐵)))
85, 6, 73bitr4i 211 1 ((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wb 104  wal 1329   = wceq 1331  wcel 1480  cdif 3063  cin 3065  c0 3358
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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-v 2683  df-dif 3068  df-in 3072  df-nul 3359
This theorem is referenced by:  disjel  3412  uneqdifeqim  3443  difprsn1  3654  diftpsn3  3656  orddif  4457  phpm  6752
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