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Theorem reldisj 3456
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 3127 . . . 4 (𝐴𝐶 ↔ ∀𝑥(𝑥𝐴𝑥𝐶))
2 pm5.44 915 . . . . . 6 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵))))
3 eldif 3121 . . . . . . 7 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
43imbi2i 225 . . . . . 6 ((𝑥𝐴𝑥 ∈ (𝐶𝐵)) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
52, 4bitr4di 197 . . . . 5 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
65sps 1524 . . . 4 (∀𝑥(𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
71, 6sylbi 120 . . 3 (𝐴𝐶 → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
87albidv 1811 . 2 (𝐴𝐶 → (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵))))
9 disj1 3455 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
10 dfss2 3127 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵)))
118, 9, 103bitr4g 222 1 (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wb 104  wal 1340   = wceq 1342  wcel 2135  cdif 3109  cin 3111  wss 3112  c0 3405
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 604  ax-in2 605  ax-io 699  ax-5 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-bndl 1496  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146
This theorem depends on definitions:  df-bi 116  df-tru 1345  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-nfc 2295  df-ral 2447  df-v 2724  df-dif 3114  df-in 3118  df-ss 3125  df-nul 3406
This theorem is referenced by:  disj2  3460  ssdifsn  3699  structcnvcnv  12364
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