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 ˙