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 )