Metamath Proof Explorer


Theorem csbeq1

Description: Analogue of dfsbcq for proper substitution into a class. (Contributed by NM, 10-Nov-2005)

Ref Expression
Assertion csbeq1 ( 𝐴 = 𝐵 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = ⦋ 𝐵 / 𝑥 ⦌ 𝐶 )

Proof

Step Hyp Ref Expression
1 dfsbcq ⊢ ( 𝐴 = 𝐵 → ( [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 ↔ [ 𝐵 / 𝑥 ] 𝑦 ∈ 𝐶 ) )
2 1 abbidv ⊢ ( 𝐴 = 𝐵 → { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } = { 𝑦 ∣ [ 𝐵 / 𝑥 ] 𝑦 ∈ 𝐶 } )
3 df-csb ⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 }
4 df-csb ⊢ ⦋ 𝐵 / 𝑥 ⦌ 𝐶 = { 𝑦 ∣ [ 𝐵 / 𝑥 ] 𝑦 ∈ 𝐶 }
5 2 3 4 3eqtr4g ⊢ ( 𝐴 = 𝐵 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = ⦋ 𝐵 / 𝑥 ⦌ 𝐶 )