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Metamath Proof Explorer

Theorem disjeq1d

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

Ref Expression
Hypothesis disjeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion disjeq1d ( 𝜑 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )

Proof

Step Hyp Ref Expression
1 disjeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 disjeq1 ⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )
3 1 2 syl ⊢ ( 𝜑 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )