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Theorem disjord 5092
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 881 . . . . . 6 (𝑎 = 𝑏 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
21a1d 26 . . . . 5 (𝑎 = 𝑏 → (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅)))
3 disjord.2 . . . . . . . . . . . 12 ((𝜑𝑥𝐴𝑥𝐵) → 𝑎 = 𝑏)
433expia 1139 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → (𝑥𝐵𝑎 = 𝑏))
54con3d 153 . . . . . . . . . 10 ((𝜑𝑥𝐴) → (¬ 𝑎 = 𝑏 → ¬ 𝑥𝐵))
65impancom 457 . . . . . . . . 9 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝑥𝐴 → ¬ 𝑥𝐵))
76ralrimiv 3153 . . . . . . . 8 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → ∀𝑥𝐴 ¬ 𝑥𝐵)
8 disj 4403 . . . . . . . 8 ((𝐴𝐵) = ∅ ↔ ∀𝑥𝐴 ¬ 𝑥𝐵)
97, 8sylibr 237 . . . . . . 7 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝐴𝐵) = ∅)
109olcd 888 . . . . . 6 ((𝜑 ∧ ¬ 𝑎 = 𝑏) → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1110expcom 419 . . . . 5 𝑎 = 𝑏 → (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅)))
122, 11pm2.61i 184 . . . 4 (𝜑 → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1312adantr 486 . . 3 ((𝜑 ∧ (𝑎𝑉𝑏𝑉)) → (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1413ralrimivva 3205 . 2 (𝜑 → ∀𝑎𝑉𝑏𝑉 (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
15 disjord.1 . . 3 (𝑎 = 𝑏𝐴 = 𝐵)
1615disjor 5085 . 2 (Disj 𝑎𝑉 𝐴 ↔ ∀𝑎𝑉𝑏𝑉 (𝑎 = 𝑏 ∨ (𝐴𝐵) = ∅))
1714, 16sylibr 237 1 (𝜑Disj 𝑎𝑉 𝐴)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3  wi 4  wa 401  wo 861  w3a 1103   = wceq 1570  wcel 2145  wral 3076  cin 3898  c0 4279  Disj wdisj 5070
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-mo 2564  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rmo 3365  df-v 3452  df-dif 3902  df-in 3906  df-nul 4280  df-disj 5071
This theorem is used by:  2wspdisj  30433
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