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Metamath Proof Explorer

Theorem elabgf

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. This version has bound-variable hypotheses in place of distinct variable restrictions. (Contributed by NM, 21-Sep-2003) (Revised by Mario Carneiro, 12-Oct-2016)

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

Proof

Step Hyp Ref Expression
1 elabgf.1 ⊢ Ⅎ 𝑥 𝐴
2 elabgf.2 ⊢ Ⅎ 𝑥 𝜓
3 elabgf.3 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
4 nfab1 ⊢ Ⅎ 𝑥 { 𝑥 ∣ 𝜑 }
5 1 4 nfel ⊢ Ⅎ 𝑥 𝐴 ∈ { 𝑥 ∣ 𝜑 }
6 5 2 nfbi ⊢ Ⅎ 𝑥 ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 )
7 eleq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } ) )
8 7 3 bibi12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 ) ↔ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) )
9 abid ⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 )
10 1 6 8 9 vtoclgf ⊢ ( 𝐴 ∈ 𝐵 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )