Metamath Proof Explorer


Theorem rngqiprnglinlem2

Description: Lemma 2 for rngqiprnglin . (Contributed by AV, 28-Feb-2025)

Ref Expression
Hypotheses rng2idlring.r ⊢ φ → R ∈ Rng
rng2idlring.i ⊢ φ → I ∈ 2Ideal ⁡ R
rng2idlring.j ⊢ J = R ↾ 𝑠 I
rng2idlring.u ⊢ φ → J ∈ Ring
rng2idlring.b ⊢ B = Base R
rng2idlring.t ⊢ · ˙ = ⋅ R
rng2idlring.1 ⊢ 1 ˙ = 1 J
rngqiprngim.g ⊢ ∼ ˙ = R ~ QG I
rngqiprngim.q ⊢ Q = R / 𝑠 ∼ ˙
Assertion rngqiprnglinlem2 ⊢ φ ∧ A ∈ B ∧ C ∈ B → A · ˙ C ∼ ˙ = A ∼ ˙ ⋅ Q C ∼ ˙

Proof

Step Hyp Ref Expression
1 rng2idlring.r ⊢ φ → R ∈ Rng
2 rng2idlring.i ⊢ φ → I ∈ 2Ideal ⁡ R
3 rng2idlring.j ⊢ J = R ↾ 𝑠 I
4 rng2idlring.u ⊢ φ → J ∈ Ring
5 rng2idlring.b ⊢ B = Base R
6 rng2idlring.t ⊢ · ˙ = ⋅ R
7 rng2idlring.1 ⊢ 1 ˙ = 1 J
8 rngqiprngim.g ⊢ ∼ ˙ = R ~ QG I
9 rngqiprngim.q ⊢ Q = R / 𝑠 ∼ ˙
10 ringrng ⊢ J ∈ Ring → J ∈ Rng
11 4 10 syl ⊢ φ → J ∈ Rng
12 3 11 eqeltrrid ⊢ φ → R ↾ 𝑠 I ∈ Rng
13 1 2 12 rng2idlsubrng ⊢ φ → I ∈ SubRng ⁡ R
14 subrngsubg ⊢ I ∈ SubRng ⁡ R → I ∈ SubGrp ⁡ R
15 13 14 syl ⊢ φ → I ∈ SubGrp ⁡ R
16 1 2 15 3jca ⊢ φ → R ∈ Rng ∧ I ∈ 2Ideal ⁡ R ∧ I ∈ SubGrp ⁡ R
17 eqid ⊢ R ~ QG I = R ~ QG I
18 8 oveq2i ⊢ R / 𝑠 ∼ ˙ = R / 𝑠 R ~ QG I
19 9 18 eqtri ⊢ Q = R / 𝑠 R ~ QG I
20 eqid ⊢ ⋅ Q = ⋅ Q
21 17 19 5 6 20 qusmulrng ⊢ R ∈ Rng ∧ I ∈ 2Ideal ⁡ R ∧ I ∈ SubGrp ⁡ R ∧ A ∈ B ∧ C ∈ B → A R ~ QG I ⋅ Q C R ~ QG I = A · ˙ C R ~ QG I
22 16 21 sylan ⊢ φ ∧ A ∈ B ∧ C ∈ B → A R ~ QG I ⋅ Q C R ~ QG I = A · ˙ C R ~ QG I
23 8 eceq2i ⊢ A ∼ ˙ = A R ~ QG I
24 8 eceq2i ⊢ C ∼ ˙ = C R ~ QG I
25 23 24 oveq12i ⊢ A ∼ ˙ ⋅ Q C ∼ ˙ = A R ~ QG I ⋅ Q C R ~ QG I
26 8 eceq2i ⊢ A · ˙ C ∼ ˙ = A · ˙ C R ~ QG I
27 22 25 26 3eqtr4g ⊢ φ ∧ A ∈ B ∧ C ∈ B → A ∼ ˙ ⋅ Q C ∼ ˙ = A · ˙ C ∼ ˙
28 27 eqcomd ⊢ φ ∧ A ∈ B ∧ C ∈ B → A · ˙ C ∼ ˙ = A ∼ ˙ ⋅ Q C ∼ ˙