Metamath Proof Explorer


Theorem subrgin

Description: The intersection of two subrings is a subring. (Contributed by Stefan O'Rear, 30-Nov-2014) (Revised by Mario Carneiro, 7-Dec-2014)

Ref Expression
Assertion subrgin ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → A ∩ B ∈ SubRing ⁡ R

Proof

Step Hyp Ref Expression
1 intprg ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → ⋂ A B = A ∩ B
2 prssi ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → A B ⊆ SubRing ⁡ R
3 prnzg ⊢ A ∈ SubRing ⁡ R → A B ≠ ∅
4 3 adantr ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → A B ≠ ∅
5 subrgint ⊢ A B ⊆ SubRing ⁡ R ∧ A B ≠ ∅ → ⋂ A B ∈ SubRing ⁡ R
6 2 4 5 syl2anc ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → ⋂ A B ∈ SubRing ⁡ R
7 1 6 eqeltrrd ⊢ A ∈ SubRing ⁡ R ∧ B ∈ SubRing ⁡ R → A ∩ B ∈ SubRing ⁡ R