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