Metamath Proof Explorer


Theorem disjeq2dv

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

Ref Expression
Hypothesis disjeq2dv.1 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐵 = 𝐶 )
Assertion disjeq2dv ( 𝜑 → ( Disj 𝑥 ∈ 𝐴 𝐵 ↔ Disj 𝑥 ∈ 𝐴 𝐶 ) )

Proof

Step Hyp Ref Expression
1 disjeq2dv.1 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐵 = 𝐶 )
2 1 ralrimiva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 𝐵 = 𝐶 )
3 disjeq2 ⊢ ( ∀ 𝑥 ∈ 𝐴 𝐵 = 𝐶 → ( Disj 𝑥 ∈ 𝐴 𝐵 ↔ Disj 𝑥 ∈ 𝐴 𝐶 ) )
4 2 3 syl ⊢ ( 𝜑 → ( Disj 𝑥 ∈ 𝐴 𝐵 ↔ Disj 𝑥 ∈ 𝐴 𝐶 ) )