Metamath Proof Explorer


Theorem elrabd

Description: Membership in a restricted class abstraction, using implicit substitution. Deduction version of elrab . (Contributed by Glauco Siliprandi, 23-Oct-2021)

Ref Expression
Hypotheses elrabd.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
elrabd.2 ⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
elrabd.3 ⊢ ( 𝜑 → 𝜒 )
Assertion elrabd ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )

Proof

Step Hyp Ref Expression
1 elrabd.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
2 elrabd.2 ⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
3 elrabd.3 ⊢ ( 𝜑 → 𝜒 )
4 1 elrab ⊢ ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ↔ ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) )
5 2 3 4 sylanbrc ⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )