Theorem sbcbidv

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

	Ref	Expression
Hypothesis	sbcbidv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	sbcbidv	⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcbidv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2		eqidd	⊢ ( 𝜑 → 𝐴 = 𝐴 )
3	2 1	sbceqbid	⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜒 ) )