Metamath Proof Explorer


Theorem rspcva

Description: Restricted specialization, using implicit substitution. (Contributed by NM, 13-Sep-2005)

Ref Expression
Hypothesis rspcv.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion rspcva ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) → 𝜓 )

Proof

Step Hyp Ref Expression
1 rspcv.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 1 rspcv ⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ∈ 𝐵 𝜑 → 𝜓 ) )
3 2 imp ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) → 𝜓 )