Metamath Proof Explorer


Theorem disjeq12i

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

Ref Expression
Hypotheses disjeq12i.1 ⊢ A = B
disjeq12i.2 ⊢ C = D
Assertion disjeq12i ⊢ Disj x ∈ A C ↔ Disj x ∈ B D

Proof

Step Hyp Ref Expression
1 disjeq12i.1 ⊢ A = B
2 disjeq12i.2 ⊢ C = D
3 disjeq2 ⊢ ∀ x ∈ A C = D → Disj x ∈ A C ↔ Disj x ∈ A D
4 2 a1i ⊢ x ∈ A → C = D
5 3 4 mprg ⊢ Disj x ∈ A C ↔ Disj x ∈ A D
6 1 disjeq1i ⊢ Disj x ∈ A D ↔ Disj x ∈ B D
7 5 6 bitri ⊢ Disj x ∈ A C ↔ Disj x ∈ B D