Theorem disjexc 32878

Description: A variant of disjex 32877, 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 883	. . 3 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
4		df-in 3920	. . . . . . 7 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)}
5	4	neeq1i 3028	. . . . . 6 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅)
6		abn0 4348	. . . . . 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 3015	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
9	8	orbi2i 925	. . 3 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)))
10		imor 866	. . 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) = ∅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   ∨ wo 860   = wceq 1567  ∃wex 1806   ∈ wcel 2149  {cab 2747   ≠ wne 2964   ∩ cin 3912  ∅c0 4294

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-10 2182  ax-11 2198  ax-12 2219  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-ne 2965  df-dif 3916  df-in 3920  df-nul 4295

This theorem is referenced by: (None)