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 φ x ψ

Proof

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