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