Metamath Proof Explorer


Theorem ssfii

Description: Any element of a set A is the intersection of a finite subset of A . (Contributed by FL, 27-Apr-2008) (Proof shortened by Mario Carneiro, 21-Mar-2015)

Ref Expression
Assertion ssfii ⊢ A ∈ V → A ⊆ fi ⁡ A

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 intsn ⊢ ⋂ x = x
3 simpl ⊢ A ∈ V ∧ x ∈ A → A ∈ V
4 simpr ⊢ A ∈ V ∧ x ∈ A → x ∈ A
5 4 snssd ⊢ A ∈ V ∧ x ∈ A → x ⊆ A
6 1 snnz ⊢ x ≠ ∅
7 6 a1i ⊢ A ∈ V ∧ x ∈ A → x ≠ ∅
8 snfi ⊢ x ∈ Fin
9 8 a1i ⊢ A ∈ V ∧ x ∈ A → x ∈ Fin
10 elfir ⊢ A ∈ V ∧ x ⊆ A ∧ x ≠ ∅ ∧ x ∈ Fin → ⋂ x ∈ fi ⁡ A
11 3 5 7 9 10 syl13anc ⊢ A ∈ V ∧ x ∈ A → ⋂ x ∈ fi ⁡ A
12 2 11 eqeltrrid ⊢ A ∈ V ∧ x ∈ A → x ∈ fi ⁡ A
13 12 ex ⊢ A ∈ V → x ∈ A → x ∈ fi ⁡ A
14 13 ssrdv ⊢ A ∈ V → A ⊆ fi ⁡ A