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Metamath Proof Explorer

Theorem abid2fOLD

Description: Obsolete version of abid2f as of 26-Feb-2025. (Contributed by NM, 5-Sep-2011) (Revised by Mario Carneiro, 7-Oct-2016) (Proof shortened by Wolf Lammen, 17-Nov-2019) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Hypothesis	abid2f.1	⊢ Ⅎ 𝑥 𝐴
	Assertion	abid2fOLD	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴

Proof

Step	Hyp	Ref	Expression
1		abid2f.1	⊢ Ⅎ 𝑥 𝐴
2		nfab1	⊢ Ⅎ 𝑥 { 𝑥 ∣ 𝑥 ∈ 𝐴 }
3	2 1	cleqf	⊢ ( { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴 ↔ ∀ 𝑥 ( 𝑥 ∈ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ 𝑥 ∈ 𝐴 ) )
4		abid	⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ 𝑥 ∈ 𝐴 )
5	3 4	mpgbir	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴