Metamath Proof Explorer


Theorem hfsshf

Description: Any subset of a hereditarily finite set is itself a hereditarily finite set. (Contributed by BTernaryTau, 17-Sep-2026)

Ref Expression
Assertion hfsshf
|- ( ( A C_ B /\ B e. HF ) -> A e. HF )

Proof

Step Hyp Ref Expression
1 hffi
 |-  ( B e. HF -> B e. Fin )
2 ssfi
 |-  ( ( B e. Fin /\ A C_ B ) -> A e. Fin )
3 2 ancoms
 |-  ( ( A C_ B /\ B e. Fin ) -> A e. Fin )
4 1 3 sylan2
 |-  ( ( A C_ B /\ B e. HF ) -> A e. Fin )
5 elhf3
 |-  ( B e. HF <-> ( B e. Fin /\ B C_ HF ) )
6 5 simprbi
 |-  ( B e. HF -> B C_ HF )
7 sstr
 |-  ( ( A C_ B /\ B C_ HF ) -> A C_ HF )
8 6 7 sylan2
 |-  ( ( A C_ B /\ B e. HF ) -> A C_ HF )
9 elhf3
 |-  ( A e. HF <-> ( A e. Fin /\ A C_ HF ) )
10 4 8 9 sylanbrc
 |-  ( ( A C_ B /\ B e. HF ) -> A e. HF )