Metamath Proof Explorer


Theorem subrngringnsg

Description: A subring is a normal subgroup. (Contributed by AV, 25-Feb-2025)

Ref Expression
Assertion subrngringnsg ⊢ A ∈ SubRng ⁡ R → A ∈ NrmSGrp ⁡ R

Proof

Step Hyp Ref Expression
1 subrngsubg ⊢ A ∈ SubRng ⁡ R → A ∈ SubGrp ⁡ R
2 subrngrcl ⊢ A ∈ SubRng ⁡ R → R ∈ Rng
3 rngabl ⊢ R ∈ Rng → R ∈ Abel
4 2 3 syl ⊢ A ∈ SubRng ⁡ R → R ∈ Abel
5 4 3anim1i ⊢ A ∈ SubRng ⁡ R ∧ x ∈ Base R ∧ y ∈ Base R → R ∈ Abel ∧ x ∈ Base R ∧ y ∈ Base R
6 5 3expb ⊢ A ∈ SubRng ⁡ R ∧ x ∈ Base R ∧ y ∈ Base R → R ∈ Abel ∧ x ∈ Base R ∧ y ∈ Base R
7 eqid ⊢ Base R = Base R
8 eqid ⊢ + R = + R
9 7 8 ablcom ⊢ R ∈ Abel ∧ x ∈ Base R ∧ y ∈ Base R → x + R y = y + R x
10 6 9 syl ⊢ A ∈ SubRng ⁡ R ∧ x ∈ Base R ∧ y ∈ Base R → x + R y = y + R x
11 10 eleq1d ⊢ A ∈ SubRng ⁡ R ∧ x ∈ Base R ∧ y ∈ Base R → x + R y ∈ A ↔ y + R x ∈ A
12 11 biimpd ⊢ A ∈ SubRng ⁡ R ∧ x ∈ Base R ∧ y ∈ Base R → x + R y ∈ A → y + R x ∈ A
13 12 ralrimivva ⊢ A ∈ SubRng ⁡ R → ∀ x ∈ Base R ∀ y ∈ Base R x + R y ∈ A → y + R x ∈ A
14 7 8 isnsg2 ⊢ A ∈ NrmSGrp ⁡ R ↔ A ∈ SubGrp ⁡ R ∧ ∀ x ∈ Base R ∀ y ∈ Base R x + R y ∈ A → y + R x ∈ A
15 1 13 14 sylanbrc ⊢ A ∈ SubRng ⁡ R → A ∈ NrmSGrp ⁡ R