Metamath Proof Explorer


Theorem ringidmlem

Description: Lemma for ringlidm and ringridm . (Contributed by FL, 18-Feb-2010) (Revised by NM, 15-Sep-2011) (Revised by Mario Carneiro, 27-Dec-2014)

Ref Expression
Hypotheses ringidm.b ⊢ B = Base R
ringidm.t ⊢ · ˙ = ⋅ R
ringidm.u ⊢ 1 ˙ = 1 R
Assertion ringidmlem ⊢ R ∈ Ring ∧ X ∈ B → 1 ˙ · ˙ X = X ∧ X · ˙ 1 ˙ = X

Proof

Step Hyp Ref Expression
1 ringidm.b ⊢ B = Base R
2 ringidm.t ⊢ · ˙ = ⋅ R
3 ringidm.u ⊢ 1 ˙ = 1 R
4 eqid ⊢ mulGrp R = mulGrp R
5 4 ringmgp ⊢ R ∈ Ring → mulGrp R ∈ Mnd
6 4 1 mgpbas ⊢ B = Base mulGrp R
7 4 2 mgpplusg ⊢ · ˙ = + mulGrp R
8 4 3 ringidval ⊢ 1 ˙ = 0 mulGrp R
9 6 7 8 mndlrid ⊢ mulGrp R ∈ Mnd ∧ X ∈ B → 1 ˙ · ˙ X = X ∧ X · ˙ 1 ˙ = X
10 5 9 sylan ⊢ R ∈ Ring ∧ X ∈ B → 1 ˙ · ˙ X = X ∧ X · ˙ 1 ˙ = X