Theorem disjex 32741

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 881	. 2 ⊢ ((𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
2		df-in 3911	. . . . . 6 ⊢ (𝐴 ∩ 𝐵) = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)}
3	2	neeq1i 3020	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ ∅ ↔ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅)
4		abn0 4337	. . . . 5 ⊢ ({𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} ≠ ∅ ↔ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
5	3, 4	bitr2i 278	. . . 4 ⊢ (∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝐴 ∩ 𝐵) ≠ ∅)
6	5	necon2bbii 3007	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))
7	6	orbi2i 923	. 2 ⊢ ((𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅) ↔ (𝐴 = 𝐵 ∨ ¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)))
8		imor 864	. 2 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (¬ ∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ∨ 𝐴 = 𝐵))
9	1, 7, 8	3bitr4ri 306	1 ⊢ ((∃𝑧(𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) → 𝐴 = 𝐵) ↔ (𝐴 = 𝐵 ∨ (𝐴 ∩ 𝐵) = ∅))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399   ∨ wo 858   = wceq 1559  ∃wex 1798   ∈ wcel 2141  {cab 2739   ≠ wne 2956   ∩ cin 3903  ∅c0 4285

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-ne 2957  df-dif 3907  df-in 3911  df-nul 4286

This theorem is referenced by: (None)