Theorem sbcthdv

Description: Deduction version of sbcth . (Contributed by NM, 30-Nov-2005) (Proof shortened by Andrew Salmon, 8-Jun-2011)

	Ref	Expression
Hypothesis	sbcthdv.1	⊢ ( 𝜑 → 𝜓 )
Assertion	sbcthdv	⊢ ( ( 𝜑 ∧ 𝐴 ∈ 𝑉 ) → [ 𝐴 / 𝑥 ] 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		sbcthdv.1	⊢ ( 𝜑 → 𝜓 )
2	1	alrimiv	⊢ ( 𝜑 → ∀ 𝑥 𝜓 )
3		spsbc	⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 𝜓 → [ 𝐴 / 𝑥 ] 𝜓 ) )
4	2 3	mpan9	⊢ ( ( 𝜑 ∧ 𝐴 ∈ 𝑉 ) → [ 𝐴 / 𝑥 ] 𝜓 )