Metamath Proof Explorer


Theorem disjnf

Description: In case x is not free in B , disjointness is not so interesting since it reduces to cases where A is a singleton. (Google Groups discussion with Peter Mazsa.) (Contributed by Thierry Arnoux, 26-Jul-2018)

Ref Expression
Assertion disjnf ⊢ Disj x ∈ A B ↔ B = ∅ ∨ ∃* x x ∈ A

Proof

Step Hyp Ref Expression
1 inidm ⊢ B ∩ B = B
2 1 eqeq1i ⊢ B ∩ B = ∅ ↔ B = ∅
3 2 orbi1i ⊢ B ∩ B = ∅ ∨ ∀ x ∈ A ∀ y ∈ A x = y ↔ B = ∅ ∨ ∀ x ∈ A ∀ y ∈ A x = y
4 eqidd ⊢ x = y → B = B
5 4 disjor ⊢ Disj x ∈ A B ↔ ∀ x ∈ A ∀ y ∈ A x = y ∨ B ∩ B = ∅
6 orcom ⊢ x = y ∨ B ∩ B = ∅ ↔ B ∩ B = ∅ ∨ x = y
7 6 ralbii ⊢ ∀ y ∈ A x = y ∨ B ∩ B = ∅ ↔ ∀ y ∈ A B ∩ B = ∅ ∨ x = y
8 r19.32v ⊢ ∀ y ∈ A B ∩ B = ∅ ∨ x = y ↔ B ∩ B = ∅ ∨ ∀ y ∈ A x = y
9 7 8 bitri ⊢ ∀ y ∈ A x = y ∨ B ∩ B = ∅ ↔ B ∩ B = ∅ ∨ ∀ y ∈ A x = y
10 9 ralbii ⊢ ∀ x ∈ A ∀ y ∈ A x = y ∨ B ∩ B = ∅ ↔ ∀ x ∈ A B ∩ B = ∅ ∨ ∀ y ∈ A x = y
11 r19.32v ⊢ ∀ x ∈ A B ∩ B = ∅ ∨ ∀ y ∈ A x = y ↔ B ∩ B = ∅ ∨ ∀ x ∈ A ∀ y ∈ A x = y
12 5 10 11 3bitri ⊢ Disj x ∈ A B ↔ B ∩ B = ∅ ∨ ∀ x ∈ A ∀ y ∈ A x = y
13 moel ⊢ ∃* x x ∈ A ↔ ∀ x ∈ A ∀ y ∈ A x = y
14 13 orbi2i ⊢ B = ∅ ∨ ∃* x x ∈ A ↔ B = ∅ ∨ ∀ x ∈ A ∀ y ∈ A x = y
15 3 12 14 3bitr4i ⊢ Disj x ∈ A B ↔ B = ∅ ∨ ∃* x x ∈ A