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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 )