Theorem sbcbid

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014)

	Ref	Expression
Hypotheses	sbcbid.1	⊢ Ⅎ 𝑥 𝜑
	sbcbid.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	sbcbid	⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcbid.1	⊢ Ⅎ 𝑥 𝜑
2		sbcbid.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3	1 2	abbid	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = { 𝑥 ∣ 𝜒 } )
4	3	eleq2d	⊢ ( 𝜑 → ( 𝐴 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝐴 ∈ { 𝑥 ∣ 𝜒 } ) )
5		df-sbc	⊢ ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜓 } )
6		df-sbc	⊢ ( [ 𝐴 / 𝑥 ] 𝜒 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜒 } )
7	4 5 6	3bitr4g	⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜒 ) )