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Theorem reldisj 3299
Description: Two ways of saying that two classes are disjoint, using the complement of 𝐵 relative to a universe 𝐶. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
reldisj (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))

Proof of Theorem reldisj
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 dfss2 2962 . . . 4 (𝐴𝐶 ↔ ∀𝑥(𝑥𝐴𝑥𝐶))
2 pm5.44 845 . . . . . 6 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵))))
3 eldif 2955 . . . . . . 7 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
43imbi2i 219 . . . . . 6 ((𝑥𝐴𝑥 ∈ (𝐶𝐵)) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
52, 4syl6bbr 191 . . . . 5 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
65sps 1446 . . . 4 (∀𝑥(𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
71, 6sylbi 118 . . 3 (𝐴𝐶 → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
87albidv 1721 . 2 (𝐴𝐶 → (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵))))
9 disj1 3298 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
10 dfss2 2962 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵)))
118, 9, 103bitr4g 216 1 (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wb 102  wal 1257   = wceq 1259  wcel 1409  cdif 2942  cin 2944  wss 2945  c0 3252
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-v 2576  df-dif 2948  df-in 2952  df-ss 2959  df-nul 3253
This theorem is referenced by:  disj2  3303
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