Metamath Proof Explorer


Theorem spesbcd

Description: form of spsbc . (Contributed by Mario Carneiro, 9-Feb-2017)

Ref Expression
Hypothesis spesbcd.1 φ[˙A/x]˙ψ
Assertion spesbcd φxψ

Proof

Step Hyp Ref Expression
1 spesbcd.1 φ[˙A/x]˙ψ
2 spesbc [˙A/x]˙ψxψ
3 1 2 syl φxψ