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Metamath Proof Explorer

Theorem elabd3

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab . (Contributed by GG, 12-Oct-2024)

Ref Expression
Hypotheses elabd3.ex ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
elabd3.is ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion elabd3 ( 𝜑 → ( 𝐴 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 elabd3.ex ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
2 elabd3.is ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
3 eqidd ⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = { 𝑥 ∣ 𝜓 } )
4 1 3 2 elabd2 ⊢ ( 𝜑 → ( 𝐴 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜒 ) )