Metamath Proof Explorer


Theorem hfelhf

Description: Any member of a hereditarily finite set is itself a hereditarily finite set. (Contributed by Scott Fenton, 16-Jul-2015) Avoid ax-reg , ax-inf2 . (Revised by BTernaryTau, 17-Sep-2026)

Ref Expression
Assertion hfelhf ( ( 𝐴 ∈ 𝐵 ∧ 𝐵 ∈ HF ) → 𝐴 ∈ HF )

Proof

Step Hyp Ref Expression
1 elhf4 ⊢ ( 𝐵 ∈ HF ↔ ( 𝐵 ∈ Fin ∧ ∀ 𝑥 ∈ 𝐵 𝑥 ∈ HF ) )
2 1 simprbi ⊢ ( 𝐵 ∈ HF → ∀ 𝑥 ∈ 𝐵 𝑥 ∈ HF )
3 eleq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∈ HF ↔ 𝐴 ∈ HF ) )
4 3 rspcva ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐵 𝑥 ∈ HF ) → 𝐴 ∈ HF )
5 2 4 sylan2 ⊢ ( ( 𝐴 ∈ 𝐵 ∧ 𝐵 ∈ HF ) → 𝐴 ∈ HF )