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 ˙