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Theorem reldisj 3271
 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 2934 . . . 4 (𝐴𝐶 ↔ ∀𝑥(𝑥𝐴𝑥𝐶))
2 pm5.44 834 . . . . . 6 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵))))
3 eldif 2927 . . . . . . 7 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
43imbi2i 215 . . . . . 6 ((𝑥𝐴𝑥 ∈ (𝐶𝐵)) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
52, 4syl6bbr 187 . . . . 5 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
65sps 1430 . . . 4 (∀𝑥(𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
71, 6sylbi 114 . . 3 (𝐴𝐶 → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
87albidv 1705 . 2 (𝐴𝐶 → (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵))))
9 disj1 3270 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
10 dfss2 2934 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵)))
118, 9, 103bitr4g 212 1 (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1241   = wceq 1243   ∈ wcel 1393   ∖ 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:  disj2  3275
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