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