Metamath Proof Explorer


Theorem disjors

Description: Two ways to say that a collection B ( i ) for i e. A is disjoint. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Assertion disjors ⊢ Disj x ∈ A B ↔ ∀ i ∈ A ∀ j ∈ A i = j ∨ ⦋ i / x⦌ B ∩ ⦋ j / x⦌ B = ∅

Proof

Step Hyp Ref Expression
1 nfcv ⊢ Ⅎ _ i B
2 nfcsb1v ⊢ Ⅎ _ x ⦋ i / x⦌ B
3 csbeq1a ⊢ x = i → B = ⦋ i / x⦌ B
4 1 2 3 cbvdisj ⊢ Disj x ∈ A B ↔ Disj i ∈ A ⦋ i / x⦌ B
5 csbeq1 ⊢ i = j → ⦋ i / x⦌ B = ⦋ j / x⦌ B
6 5 disjor ⊢ Disj i ∈ A ⦋ i / x⦌ B ↔ ∀ i ∈ A ∀ j ∈ A i = j ∨ ⦋ i / x⦌ B ∩ ⦋ j / x⦌ B = ∅
7 4 6 bitri ⊢ Disj x ∈ A B ↔ ∀ i ∈ A ∀ j ∈ A i = j ∨ ⦋ i / x⦌ B ∩ ⦋ j / x⦌ B = ∅