Metamath Proof Explorer


Theorem istdrg

Description: Express the predicate " R is a topological ring". (Contributed by Mario Carneiro, 5-Oct-2015)

Ref Expression
Hypotheses istrg.1 ⊢ M = mulGrp R
istdrg.1 ⊢ U = Unit ⁡ R
Assertion istdrg ⊢ R ∈ TopDRing ↔ R ∈ TopRing ∧ R ∈ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp

Proof

Step Hyp Ref Expression
1 istrg.1 ⊢ M = mulGrp R
2 istdrg.1 ⊢ U = Unit ⁡ R
3 elin ⊢ R ∈ TopRing ∩ DivRing ↔ R ∈ TopRing ∧ R ∈ DivRing
4 3 anbi1i ⊢ R ∈ TopRing ∩ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp ↔ R ∈ TopRing ∧ R ∈ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp
5 fveq2 ⊢ r = R → mulGrp r = mulGrp R
6 5 1 eqtr4di ⊢ r = R → mulGrp r = M
7 fveq2 ⊢ r = R → Unit ⁡ r = Unit ⁡ R
8 7 2 eqtr4di ⊢ r = R → Unit ⁡ r = U
9 6 8 oveq12d ⊢ r = R → mulGrp r ↾ 𝑠 Unit ⁡ r = M ↾ 𝑠 U
10 9 eleq1d ⊢ r = R → mulGrp r ↾ 𝑠 Unit ⁡ r ∈ TopGrp ↔ M ↾ 𝑠 U ∈ TopGrp
11 df-tdrg ⊢ TopDRing = r ∈ TopRing ∩ DivRing | mulGrp r ↾ 𝑠 Unit ⁡ r ∈ TopGrp
12 10 11 elrab2 ⊢ R ∈ TopDRing ↔ R ∈ TopRing ∩ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp
13 df-3an ⊢ R ∈ TopRing ∧ R ∈ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp ↔ R ∈ TopRing ∧ R ∈ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp
14 4 12 13 3bitr4i ⊢ R ∈ TopDRing ↔ R ∈ TopRing ∧ R ∈ DivRing ∧ M ↾ 𝑠 U ∈ TopGrp