Metamath Proof Explorer


Theorem subrngin

Description: The intersection of two subrings is a subring. (Contributed by AV, 15-Feb-2025)

Ref Expression
Assertion subrngin ⊢ A ∈ SubRng ⁡ R ∧ B ∈ SubRng ⁡ R → A ∩ B ∈ SubRng ⁡ R

Proof

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