Metamath Proof Explorer


Theorem disjex

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

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

Proof

Step Hyp Ref Expression
1 orcom ⊢ A = B ∨ ¬ ∃ z z ∈ A ∧ z ∈ B ↔ ¬ ∃ z z ∈ A ∧ z ∈ B ∨ A = B
2 df-in ⊢ A ∩ B = z | z ∈ A ∧ z ∈ B
3 2 neeq1i ⊢ A ∩ B ≠ ∅ ↔ z | z ∈ A ∧ z ∈ B ≠ ∅
4 abn0 ⊢ z | z ∈ A ∧ z ∈ B ≠ ∅ ↔ ∃ z z ∈ A ∧ z ∈ B
5 3 4 bitr2i ⊢ ∃ z z ∈ A ∧ z ∈ B ↔ A ∩ B ≠ ∅
6 5 necon2bbii ⊢ A ∩ B = ∅ ↔ ¬ ∃ z z ∈ A ∧ z ∈ B
7 6 orbi2i ⊢ A = B ∨ A ∩ B = ∅ ↔ A = B ∨ ¬ ∃ z z ∈ A ∧ z ∈ B
8 imor ⊢ ∃ z z ∈ A ∧ z ∈ B → A = B ↔ ¬ ∃ z z ∈ A ∧ z ∈ B ∨ A = B
9 1 7 8 3bitr4ri ⊢ ∃ z z ∈ A ∧ z ∈ B → A = B ↔ A = B ∨ A ∩ B = ∅