Metamath Proof Explorer


Theorem rspccv

Description: Restricted specialization, using implicit substitution. (Contributed by NM, 2-Feb-2006)

Ref Expression
Hypothesis rspcv.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion rspccv ( ∀ 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 com12 ⊢ ( ∀ 𝑥 ∈ 𝐵 𝜑 → ( 𝐴 ∈ 𝐵 → 𝜓 ) )