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Theorem disjord 5028
Description: Conditions for a collection of sets 𝐴(𝑎) for 𝑎𝑉 to be disjoint. (Contributed by AV, 9-Jan-2022.)
Hypotheses
Ref Expression
disjord.1 (𝑎 = 𝑏𝐴 = 𝐵)
disjord.2 ((𝜑𝑥𝐴𝑥𝐵) → 𝑎 = 𝑏)
Assertion
Ref Expression
disjord (𝜑Disj 𝑎𝑉 𝐴)
Distinct variable groups:   𝐴,𝑏,𝑥   𝐵,𝑎,𝑥   𝑉,𝑎,𝑏,𝑥   𝜑,𝑎,𝑏,𝑥
Allowed substitution hints:   𝐴(𝑎)   𝐵(𝑏)

Proof of Theorem disjord
StepHypRef Expression
1 orc 866 . . . . . 6 (𝑎 = 𝑏 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
21a1d 25 . . . . 5 (𝑎 = 𝑏 → (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅)))
3 disjord.2 . . . . . . . . . . . 12 ((𝜑𝑥𝐴𝑥𝐵) → 𝑎 = 𝑏)
433expia 1122 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → (𝑥𝐵𝑎 = 𝑏))
54con3d 155 . . . . . . . . . 10 ((𝜑𝑥𝐴) → (¬ 𝑎 = 𝑏 → ¬ 𝑥𝐵))
65impancom 455 . . . . . . . . 9 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝑥𝐴 → ¬ 𝑥𝐵))
76ralrimiv 3096 . . . . . . . 8 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → ∀𝑥𝐴 ¬ 𝑥𝐵)
8 disj 4347 . . . . . . . 8 ((𝐴𝐵) = ∅ ↔ ∀𝑥𝐴 ¬ 𝑥𝐵)
97, 8sylibr 237 . . . . . . 7 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝐴𝐵) = ∅)
109olcd 873 . . . . . 6 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1110expcom 417 . . . . 5 𝑎 = 𝑏 → (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅)))
122, 11pm2.61i 185 . . . 4 (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1312adantr 484 . . 3 ((𝜑 ∧ (𝑎𝑉𝑏𝑉)) → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1413ralrimivva 3104 . 2 (𝜑 → ∀𝑎𝑉𝑏𝑉 (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
15 disjord.1 . . 3 (𝑎 = 𝑏𝐴 = 𝐵)
1615disjor 5020 . 2 (Disj 𝑎𝑉 𝐴 ↔ ∀𝑎𝑉𝑏𝑉 (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1714, 16sylibr 237 1 (𝜑Disj 𝑎𝑉 𝐴)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 399  wo 846  w3a 1088   = wceq 1542  wcel 2114  wral 3054  cin 3852  c0 4221  Disj wdisj 5005
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-11 2162  ax-ext 2711
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1787  df-sb 2075  df-mo 2541  df-clab 2718  df-cleq 2731  df-clel 2812  df-ral 3059  df-rmo 3062  df-v 3402  df-dif 3856  df-in 3860  df-nul 4222  df-disj 5006
This theorem is referenced by:  2wspdisj  27912
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