Theorem csbeq2

Description: Substituting into equivalent classes gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018)

	Ref	Expression
Assertion	csbeq2	⊢ ( ∀ 𝑥 𝐵 = 𝐶 → ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 )

Proof

Step	Hyp	Ref	Expression
1		eleq2	⊢ ( 𝐵 = 𝐶 → ( 𝑦 ∈ 𝐵 ↔ 𝑦 ∈ 𝐶 ) )
2	1	alimi	⊢ ( ∀ 𝑥 𝐵 = 𝐶 → ∀ 𝑥 ( 𝑦 ∈ 𝐵 ↔ 𝑦 ∈ 𝐶 ) )
3		sbcbi2	⊢ ( ∀ 𝑥 ( 𝑦 ∈ 𝐵 ↔ 𝑦 ∈ 𝐶 ) → ( [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 ↔ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 ) )
4	2 3	syl	⊢ ( ∀ 𝑥 𝐵 = 𝐶 → ( [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 ↔ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 ) )
5	4	abbidv	⊢ ( ∀ 𝑥 𝐵 = 𝐶 → { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } )
6		df-csb	⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 }
7		df-csb	⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 }
8	5 6 7	3eqtr4g	⊢ ( ∀ 𝑥 𝐵 = 𝐶 → ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 )