Metamath Proof Explorer


Theorem ringidss

Description: A subset of the multiplicative group has the multiplicative identity as its identity if the identity is in the subset. (Contributed by Mario Carneiro, 27-Dec-2014) (Revised by Mario Carneiro, 30-Apr-2015)

Ref Expression
Hypotheses ringidss.g M = mulGrp R 𝑠 A
ringidss.b B = Base R
ringidss.u 1 ˙ = 1 R
Assertion ringidss R Ring A B 1 ˙ A 1 ˙ = 0 M

Proof

Step Hyp Ref Expression
1 ringidss.g M = mulGrp R 𝑠 A
2 ringidss.b B = Base R
3 ringidss.u 1 ˙ = 1 R
4 eqid Base M = Base M
5 eqid 0 M = 0 M
6 eqid + M = + M
7 simp3 R Ring A B 1 ˙ A 1 ˙ A
8 eqid mulGrp R = mulGrp R
9 8 2 mgpbas B = Base mulGrp R
10 1 9 ressbas2 A B A = Base M
11 10 3ad2ant2 R Ring A B 1 ˙ A A = Base M
12 7 11 eleqtrd R Ring A B 1 ˙ A 1 ˙ Base M
13 simp2 R Ring A B 1 ˙ A A B
14 11 13 eqsstrrd R Ring A B 1 ˙ A Base M B
15 14 sselda R Ring A B 1 ˙ A y Base M y B
16 fvex Base M V
17 11 16 eqeltrdi R Ring A B 1 ˙ A A V
18 eqid R = R
19 8 18 mgpplusg R = + mulGrp R
20 1 19 ressplusg A V R = + M
21 17 20 syl R Ring A B 1 ˙ A R = + M
22 21 adantr R Ring A B 1 ˙ A y B R = + M
23 22 oveqd R Ring A B 1 ˙ A y B 1 ˙ R y = 1 ˙ + M y
24 2 18 3 ringlidm R Ring y B 1 ˙ R y = y
25 24 3ad2antl1 R Ring A B 1 ˙ A y B 1 ˙ R y = y
26 23 25 eqtr3d R Ring A B 1 ˙ A y B 1 ˙ + M y = y
27 15 26 syldan R Ring A B 1 ˙ A y Base M 1 ˙ + M y = y
28 22 oveqd R Ring A B 1 ˙ A y B y R 1 ˙ = y + M 1 ˙
29 2 18 3 ringridm R Ring y B y R 1 ˙ = y
30 29 3ad2antl1 R Ring A B 1 ˙ A y B y R 1 ˙ = y
31 28 30 eqtr3d R Ring A B 1 ˙ A y B y + M 1 ˙ = y
32 15 31 syldan R Ring A B 1 ˙ A y Base M y + M 1 ˙ = y
33 4 5 6 12 27 32 ismgmid2 R Ring A B 1 ˙ A 1 ˙ = 0 M