Metamath Proof Explorer


Theorem ress1r

Description: 1r is unaffected by restriction. This is a bit more generic than subrg1 . (Contributed by Thierry Arnoux, 6-Sep-2018)

Ref Expression
Hypotheses ress1r.s ⊢ S = R ↾ 𝑠 A
ress1r.b ⊢ B = Base R
ress1r.1 ⊢ 1 ˙ = 1 R
Assertion ress1r ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → 1 ˙ = 1 S

Proof

Step Hyp Ref Expression
1 ress1r.s ⊢ S = R ↾ 𝑠 A
2 ress1r.b ⊢ B = Base R
3 ress1r.1 ⊢ 1 ˙ = 1 R
4 1 2 ressbas2 ⊢ A ⊆ B → A = Base S
5 4 3ad2ant3 ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → A = Base S
6 simp3 ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → A ⊆ B
7 2 fvexi ⊢ B ∈ V
8 ssexg ⊢ A ⊆ B ∧ B ∈ V → A ∈ V
9 6 7 8 sylancl ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → A ∈ V
10 eqid ⊢ ⋅ R = ⋅ R
11 1 10 ressmulr ⊢ A ∈ V → ⋅ R = ⋅ S
12 9 11 syl ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → ⋅ R = ⋅ S
13 simp2 ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → 1 ˙ ∈ A
14 simpl1 ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → R ∈ Ring
15 6 sselda ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → x ∈ B
16 2 10 3 ringlidm ⊢ R ∈ Ring ∧ x ∈ B → 1 ˙ ⋅ R x = x
17 14 15 16 syl2anc ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → 1 ˙ ⋅ R x = x
18 2 10 3 ringridm ⊢ R ∈ Ring ∧ x ∈ B → x ⋅ R 1 ˙ = x
19 14 15 18 syl2anc ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → x ⋅ R 1 ˙ = x
20 5 12 13 17 19 ringurd ⊢ R ∈ Ring ∧ 1 ˙ ∈ A ∧ A ⊆ B → 1 ˙ = 1 S