Metamath Proof Explorer


Theorem disjeq1

Description: Equality theorem for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Assertion disjeq1 ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
2 disjss1 ⊢ ( 𝐵 ⊆ 𝐴 → ( Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶 ) )
3 1 2 syl ⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶 ) )
4 eqimss ⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
5 disjss1 ⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )
6 4 5 syl ⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )
7 3 6 impbid ⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )