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Metamath Proof Explorer

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 )