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×𝑠ring
pzriprng.i I=×0
pzriprng.j J=R𝑠I
pzriprng.1 1˙=1J
pzriprng.g ˙=R~QGI
pzriprng.q Q=R/𝑠˙
Assertion pzriprnglem13 QRing

Proof

Step Hyp Ref Expression
1 pzriprng.r R=ring×𝑠ring
2 pzriprng.i I=×0
3 pzriprng.j J=R𝑠I
4 pzriprng.1 1˙=1J
5 pzriprng.g ˙=R~QGI
6 pzriprng.q Q=R/𝑠˙
7 1 pzriprnglem1 RRng
8 1 2 3 pzriprnglem8 I2IdealR
9 1 2 pzriprnglem4 ISubGrpR
10 5 oveq2i R/𝑠˙=R/𝑠R~QGI
11 6 10 eqtri Q=R/𝑠R~QGI
12 eqid 2IdealR=2IdealR
13 11 12 qus2idrng RRngI2IdealRISubGrpRQRng
14 7 8 9 13 mp3an QRng
15 1z 1
16 zex V
17 snex 1V
18 16 17 xpex ×1V
19 18 snid ×1×1
20 sneq y=1y=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 1×1×1y×1×y
25 15 19 24 mp2an y×1×y
26 eliun ×1y×yy×1×y
27 25 26 mpbir ×1y×y
28 1 2 3 4 5 6 pzriprnglem11 BaseQ=y×y
29 27 28 eleqtrri ×1BaseQ
30 oveq1 i=×1iQx=×1Qx
31 30 eqeq1d i=×1iQx=x×1Qx=x
32 31 ovanraleqv i=×1xBaseQiQx=xxQi=xxBaseQ×1Qx=xxQ×1=x
33 id ×1BaseQ×1BaseQ
34 1 2 3 4 5 6 pzriprnglem12 xBaseQ×1Qx=xxQ×1=x
35 34 a1i ×1BaseQxBaseQ×1Qx=xxQ×1=x
36 35 ralrimiv ×1BaseQxBaseQ×1Qx=xxQ×1=x
37 32 33 36 rspcedvdw ×1BaseQiBaseQxBaseQiQx=xxQi=x
38 29 37 ax-mp iBaseQxBaseQiQx=xxQi=x
39 eqid BaseQ=BaseQ
40 eqid Q=Q
41 39 40 isringrng QRingQRngiBaseQxBaseQiQx=xxQi=x
42 14 38 41 mpbir2an QRing