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Theorem disjord 5018
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 864 . . . . . 6 (𝑎 = 𝑏 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
21a1d 25 . . . . 5 (𝑎 = 𝑏 → (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅)))
3 disjord.2 . . . . . . . . . . . 12 ((𝜑𝑥𝐴𝑥𝐵) → 𝑎 = 𝑏)
433expia 1118 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → (𝑥𝐵𝑎 = 𝑏))
54con3d 155 . . . . . . . . . 10 ((𝜑𝑥𝐴) → (¬ 𝑎 = 𝑏 → ¬ 𝑥𝐵))
65impancom 455 . . . . . . . . 9 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝑥𝐴 → ¬ 𝑥𝐵))
76ralrimiv 3148 . . . . . . . 8 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → ∀𝑥𝐴 ¬ 𝑥𝐵)
8 disj 4355 . . . . . . . 8 ((𝐴𝐵) = ∅ ↔ ∀𝑥𝐴 ¬ 𝑥𝐵)
97, 8sylibr 237 . . . . . . 7 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝐴𝐵) = ∅)
109olcd 871 . . . . . 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 3156 . 2 (𝜑 → ∀𝑎𝑉𝑏𝑉 (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
15 disjord.1 . . 3 (𝑎 = 𝑏𝐴 = 𝐵)
1615disjor 5010 . 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 844  w3a 1084   = wceq 1538  wcel 2111  wral 3106  cin 3880  c0 4243  Disj wdisj 4995
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-11 2158  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-ex 1782  df-sb 2070  df-mo 2598  df-clab 2777  df-cleq 2791  df-clel 2870  df-ral 3111  df-rmo 3114  df-v 3443  df-dif 3884  df-in 3888  df-nul 4244  df-disj 4996
This theorem is referenced by:  2wspdisj  27748
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