Metamath Proof Explorer


Theorem disjeq1i

Description: Equality theorem for disjoint collection. Inference version. (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypothesis disjeq1i.1 ⊢ A = B
Assertion disjeq1i ⊢ Disj x ∈ A C ↔ Disj x ∈ B C

Proof

Step Hyp Ref Expression
1 disjeq1i.1 ⊢ A = B
2 1 rmoeqi ⊢ ∃* x ∈ A t ∈ C ↔ ∃* x ∈ B t ∈ C
3 2 albii ⊢ ∀ t ∃* x ∈ A t ∈ C ↔ ∀ t ∃* x ∈ B t ∈ C
4 df-disj ⊢ Disj x ∈ A C ↔ ∀ t ∃* x ∈ A t ∈ C
5 df-disj ⊢ Disj x ∈ B C ↔ ∀ t ∃* x ∈ B t ∈ C
6 3 4 5 3bitr4i ⊢ Disj x ∈ A C ↔ Disj x ∈ B C