Metamath Proof Explorer


Theorem rspccva

Description: Restricted specialization, using implicit substitution. (Contributed by NM, 26-Jul-2006) (Proof shortened by Andrew Salmon, 8-Jun-2011)

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

Proof

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