Theorem disjex 32605

Description: Two ways to say that two classes are disjoint (or equal). (Contributed by Thierry Arnoux, 4-Oct-2016.)

Assertion

Ref	Expression
disjex	⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))

Distinct variable groups:   𝑧,𝐴   𝑧,𝐵

Proof of Theorem disjex

Step	Hyp	Ref	Expression
1		orcom 871	. 2 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
2		df-in 3958	. . . . . 6 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)}
3	2	neeq1i 3005	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅)
4		abn0 4385	. . . . 5 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
5	3, 4	bitr2i 276	. . . 4 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅)
6	5	necon2bbii 2992	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
7	6	orbi2i 913	. 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)))
8		imor 854	. 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
9	1, 7, 8	3bitr4ri 304	1 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 848   = wceq 1540  ∃wex 1779   ∈ wcel 2108  {cab 2714   ≠ wne 2940   ∩ cin 3950  ∅c0 4333

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-ne 2941  df-dif 3954  df-in 3958  df-nul 4334

This theorem is referenced by: (None)