Metamath Proof Explorer


Theorem elrabsf

Description: Membership in a restricted class abstraction, expressed with explicit class substitution. (The variation elrabf has implicit substitution). The hypothesis specifies that x must not be a free variable in B . (Contributed by NM, 30-Sep-2003) (Proof shortened by Mario Carneiro, 13-Oct-2016)

Ref Expression
Hypothesis elrabsf.1 ⊢ Ⅎ 𝑥 𝐵
Assertion elrabsf ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ( 𝐴 ∈ 𝐵 ∧ [ 𝐴 / 𝑥 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 elrabsf.1 ⊢ Ⅎ 𝑥 𝐵
2 dfsbcq ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
3 nfcv ⊢ Ⅎ 𝑦 𝐵
4 nfv ⊢ Ⅎ 𝑦 𝜑
5 nfsbc1v ⊢ Ⅎ 𝑥 [ 𝑦 / 𝑥 ] 𝜑
6 sbceq1a ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) )
7 1 3 4 5 6 cbvrabw ⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑦 ∈ 𝐵 ∣ [ 𝑦 / 𝑥 ] 𝜑 }
8 2 7 elrab2 ⊢ ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ( 𝐴 ∈ 𝐵 ∧ [ 𝐴 / 𝑥 ] 𝜑 ) )