Metamath Proof Explorer


Theorem pzriprnglem13

Description: Lemma 13 for pzriprng : Q is a unital ring. (Contributed by AV, 23-Mar-2025)

Ref Expression
Hypotheses pzriprng.r R = ring × 𝑠 ring
pzriprng.i I = × 0
pzriprng.j J = R 𝑠 I
pzriprng.1 1 ˙ = 1 J
pzriprng.g ˙ = R ~ QG I
pzriprng.q Q = R / 𝑠 ˙
Assertion pzriprnglem13 Q Ring

Proof

Step Hyp Ref Expression
1 pzriprng.r R = ring × 𝑠 ring
2 pzriprng.i I = × 0
3 pzriprng.j J = R 𝑠 I
4 pzriprng.1 1 ˙ = 1 J
5 pzriprng.g ˙ = R ~ QG I
6 pzriprng.q Q = R / 𝑠 ˙
7 1 pzriprnglem1 R Rng
8 1 2 3 pzriprnglem8 I 2Ideal R
9 1 2 pzriprnglem4 I SubGrp R
10 5 oveq2i R / 𝑠 ˙ = R / 𝑠 R ~ QG I
11 6 10 eqtri Q = R / 𝑠 R ~ QG I
12 eqid 2Ideal R = 2Ideal R
13 11 12 qus2idrng R Rng I 2Ideal R I SubGrp R Q Rng
14 7 8 9 13 mp3an Q Rng
15 1z 1
16 zex V
17 snex 1 V
18 16 17 xpex × 1 V
19 18 snid × 1 × 1
20 sneq y = 1 y = 1
21 20 xpeq2d y = 1 × y = × 1
22 21 sneqd y = 1 × y = × 1
23 22 eleq2d y = 1 × 1 × y × 1 × 1
24 23 rspcev 1 × 1 × 1 y × 1 × y
25 15 19 24 mp2an y × 1 × y
26 eliun × 1 y × y y × 1 × y
27 25 26 mpbir × 1 y × y
28 1 2 3 4 5 6 pzriprnglem11 Base Q = y × y
29 27 28 eleqtrri × 1 Base Q
30 oveq1 i = × 1 i Q x = × 1 Q x
31 30 eqeq1d i = × 1 i Q x = x × 1 Q x = x
32 31 ovanraleqv i = × 1 x Base Q i Q x = x x Q i = x x Base Q × 1 Q x = x x Q × 1 = x
33 id × 1 Base Q × 1 Base Q
34 1 2 3 4 5 6 pzriprnglem12 x Base Q × 1 Q x = x x Q × 1 = x
35 34 a1i × 1 Base Q x Base Q × 1 Q x = x x Q × 1 = x
36 35 ralrimiv × 1 Base Q x Base Q × 1 Q x = x x Q × 1 = x
37 32 33 36 rspcedvdw × 1 Base Q i Base Q x Base Q i Q x = x x Q i = x
38 29 37 ax-mp i Base Q x Base Q i Q x = x x Q i = x
39 eqid Base Q = Base Q
40 eqid Q = Q
41 39 40 isringrng Q Ring Q Rng i Base Q x Base Q i Q x = x x Q i = x
42 14 38 41 mpbir2an Q Ring