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Metamath Proof Explorer

Theorem elabg

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. (Contributed by NM, 14-Apr-1995) Avoid ax-13 . (Revised by SN, 23-Nov-2022) Avoid ax-10 , ax-11 , ax-12 . (Revised by SN, 5-Oct-2024)

		Ref	Expression
	Hypothesis	elabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
	Assertion	elabg	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		elabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2	1	ax-gen	⊢ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3		elabgt	⊢ ( ( 𝐴 ∈ 𝑉 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )
4	2 3	mpan2	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )