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Theorem reldisjOLD 4352
 Description: Obsolete version of reldisj 4351 as of 28-Jun-2024. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
reldisjOLD (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))

Proof of Theorem reldisjOLD
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 dfss2 3880 . . . 4 (𝐴𝐶 ↔ ∀𝑥(𝑥𝐴𝑥𝐶))
2 pm5.44 546 . . . . . 6 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵))))
3 eldif 3870 . . . . . . 7 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
43imbi2i 339 . . . . . 6 ((𝑥𝐴𝑥 ∈ (𝐶𝐵)) ↔ (𝑥𝐴 → (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
52, 4bitr4di 292 . . . . 5 ((𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
65sps 2182 . . . 4 (∀𝑥(𝑥𝐴𝑥𝐶) → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
71, 6sylbi 220 . . 3 (𝐴𝐶 → ((𝑥𝐴 → ¬ 𝑥𝐵) ↔ (𝑥𝐴𝑥 ∈ (𝐶𝐵))))
87albidv 1921 . 2 (𝐴𝐶 → (∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵))))
9 disj1 4350 . 2 ((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))
10 dfss2 3880 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ ∀𝑥(𝑥𝐴𝑥 ∈ (𝐶𝐵)))
118, 9, 103bitr4g 317 1 (𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536   = wceq 1538   ∈ wcel 2111   ∖ cdif 3857   ∩ cin 3859   ⊆ wss 3860  ∅c0 4227 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-12 2175  ax-ext 2729 This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-ral 3075  df-v 3411  df-dif 3863  df-in 3867  df-ss 3877  df-nul 4228 This theorem is referenced by: (None)
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