Theorem disjexc 33009

Description: A variant of disjex 33008, applicable for more generic families. (Contributed by Thierry Arnoux, 4-Oct-2016.)

Hypothesis

Ref	Expression
disjexc.1	⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)

Assertion

Ref	Expression
disjexc	⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))

Distinct variable groups:   𝑧,𝐴   𝑧,𝐵

Allowed substitution hints:   𝐴(𝑥, 𝑦)   𝐵(𝑥, 𝑦)

Proof of Theorem disjexc

Step	Hyp	Ref	Expression
1		disjexc.1	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
2	1	imim2i 17	. 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵))
3		orcom 884	. . 3 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
4		df-in 3913	. . . . . . 7 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)}
5	4	neeq1i 3024	. . . . . 6 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅)
6		abn0 4341	. . . . . 6 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
7	5, 6	bitr2i 279	. . . . 5 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅)
8	7	necon2bbii 3011	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
9	8	orbi2i 926	. . 3 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)))
10		imor 867	. . 3 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
11	3, 9, 10	3bitr4i 306	. 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵))
12	2, 11	sylibr 237	1 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝑥 = 𝑦) → (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1570  ∃wex 1812   ∈ wcel 2146  {cab 2743   ≠ wne 2960   ∩ cin 3905  ∅c0 4286

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-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2744  df-cleq 2757  df-ne 2961  df-dif 3909  df-in 3913  df-nul 4287

This theorem is used by: (None)