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Metamath Proof Explorer

Theorem elab2g

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 13-Sep-1995)

Ref Expression
Hypotheses elab2g.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
elab2g.2 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
Assertion elab2g ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elab2g.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 elab2g.2 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
3 2 eleq2i ⊢ ( 𝐴 ∈ 𝐵 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
4 1 elabg ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )
5 3 4 bitrid ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) )