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 )