Metamath Proof Explorer


Theorem disjexc

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

Ref Expression
Hypothesis disjexc.1 ⊢ x = y → A = B
Assertion disjexc ⊢ ∃ z z ∈ A ∧ z ∈ B → x = y → A = B ∨ A ∩ B = ∅

Proof

Step Hyp Ref Expression
1 disjexc.1 ⊢ x = y → A = B
2 1 imim2i ⊢ ∃ z z ∈ A ∧ z ∈ B → x = y → ∃ z z ∈ A ∧ z ∈ B → A = B
3 orcom ⊢ A = B ∨ ¬ ∃ z z ∈ A ∧ z ∈ B ↔ ¬ ∃ z z ∈ A ∧ z ∈ B ∨ A = B
4 df-in ⊢ A ∩ B = z | z ∈ A ∧ z ∈ B
5 4 neeq1i ⊢ A ∩ B ≠ ∅ ↔ z | z ∈ A ∧ z ∈ B ≠ ∅
6 abn0 ⊢ z | z ∈ A ∧ z ∈ B ≠ ∅ ↔ ∃ z z ∈ A ∧ z ∈ B
7 5 6 bitr2i ⊢ ∃ z z ∈ A ∧ z ∈ B ↔ A ∩ B ≠ ∅
8 7 necon2bbii ⊢ A ∩ B = ∅ ↔ ¬ ∃ z z ∈ A ∧ z ∈ B
9 8 orbi2i ⊢ A = B ∨ A ∩ B = ∅ ↔ A = B ∨ ¬ ∃ z z ∈ A ∧ z ∈ B
10 imor ⊢ ∃ z z ∈ A ∧ z ∈ B → A = B ↔ ¬ ∃ z z ∈ A ∧ z ∈ B ∨ A = B
11 3 9 10 3bitr4i ⊢ A = B ∨ A ∩ B = ∅ ↔ ∃ z z ∈ A ∧ z ∈ B → A = B
12 2 11 sylibr ⊢ ∃ z z ∈ A ∧ z ∈ B → x = y → A = B ∨ A ∩ B = ∅