Metamath Proof Explorer


Theorem sbcbidv

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014) Drop ax-12 . (Revised by Gino Giotto, 1-Dec-2023)

Ref Expression
Hypothesis sbcbidv.1 φψχ
Assertion sbcbidv φ[˙A/x]˙ψ[˙A/x]˙χ

Proof

Step Hyp Ref Expression
1 sbcbidv.1 φψχ
2 eqidd φA=A
3 2 1 sbceqbid φ[˙A/x]˙ψ[˙A/x]˙χ