Metamath Proof Explorer


Theorem subrgdvds

Description: If an element divides another in a subring, then it also divides the other in the parent ring. (Contributed by Mario Carneiro, 4-Dec-2014)

Ref Expression
Hypotheses subrgdvds.1 ⊢ S = R ↾ 𝑠 A
subrgdvds.2 ⊢ ∥ ˙ = ∥ r ⁡ R
subrgdvds.3 ⊢ E = ∥ r ⁡ S
Assertion subrgdvds ⊢ A ∈ SubRing ⁡ R → E ⊆ ∥ ˙

Proof

Step Hyp Ref Expression
1 subrgdvds.1 ⊢ S = R ↾ 𝑠 A
2 subrgdvds.2 ⊢ ∥ ˙ = ∥ r ⁡ R
3 subrgdvds.3 ⊢ E = ∥ r ⁡ S
4 3 reldvdsr ⊢ Rel ⁡ E
5 4 a1i ⊢ A ∈ SubRing ⁡ R → Rel ⁡ E
6 1 subrgbas ⊢ A ∈ SubRing ⁡ R → A = Base S
7 eqid ⊢ Base R = Base R
8 7 subrgss ⊢ A ∈ SubRing ⁡ R → A ⊆ Base R
9 6 8 eqsstrrd ⊢ A ∈ SubRing ⁡ R → Base S ⊆ Base R
10 9 sseld ⊢ A ∈ SubRing ⁡ R → x ∈ Base S → x ∈ Base R
11 eqid ⊢ ⋅ R = ⋅ R
12 1 11 ressmulr ⊢ A ∈ SubRing ⁡ R → ⋅ R = ⋅ S
13 12 oveqd ⊢ A ∈ SubRing ⁡ R → z ⋅ R x = z ⋅ S x
14 13 eqeq1d ⊢ A ∈ SubRing ⁡ R → z ⋅ R x = y ↔ z ⋅ S x = y
15 14 rexbidv ⊢ A ∈ SubRing ⁡ R → ∃ z ∈ Base S z ⋅ R x = y ↔ ∃ z ∈ Base S z ⋅ S x = y
16 ssrexv ⊢ Base S ⊆ Base R → ∃ z ∈ Base S z ⋅ R x = y → ∃ z ∈ Base R z ⋅ R x = y
17 9 16 syl ⊢ A ∈ SubRing ⁡ R → ∃ z ∈ Base S z ⋅ R x = y → ∃ z ∈ Base R z ⋅ R x = y
18 15 17 sylbird ⊢ A ∈ SubRing ⁡ R → ∃ z ∈ Base S z ⋅ S x = y → ∃ z ∈ Base R z ⋅ R x = y
19 10 18 anim12d ⊢ A ∈ SubRing ⁡ R → x ∈ Base S ∧ ∃ z ∈ Base S z ⋅ S x = y → x ∈ Base R ∧ ∃ z ∈ Base R z ⋅ R x = y
20 eqid ⊢ Base S = Base S
21 eqid ⊢ ⋅ S = ⋅ S
22 20 3 21 dvdsr ⊢ x E y ↔ x ∈ Base S ∧ ∃ z ∈ Base S z ⋅ S x = y
23 7 2 11 dvdsr ⊢ x ∥ ˙ y ↔ x ∈ Base R ∧ ∃ z ∈ Base R z ⋅ R x = y
24 19 22 23 3imtr4g ⊢ A ∈ SubRing ⁡ R → x E y → x ∥ ˙ y
25 df-br ⊢ x E y ↔ x y ∈ E
26 df-br ⊢ x ∥ ˙ y ↔ x y ∈ ∥ ˙
27 24 25 26 3imtr3g ⊢ A ∈ SubRing ⁡ R → x y ∈ E → x y ∈ ∥ ˙
28 5 27 relssdv ⊢ A ∈ SubRing ⁡ R → E ⊆ ∥ ˙