Metamath Proof Explorer


Theorem ringidmlem

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

Ref Expression
Hypotheses rngidm.b B = Base R
rngidm.t · ˙ = R
rngidm.u 1 ˙ = 1 R
Assertion ringidmlem R Ring X B 1 ˙ · ˙ X = X X · ˙ 1 ˙ = X

Proof

Step Hyp Ref Expression
1 rngidm.b B = Base R
2 rngidm.t · ˙ = R
3 rngidm.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