Metamath Proof Explorer


Theorem pzriprnglem4

Description: Lemma 4 for pzriprng : I is a subgroup of R . (Contributed by AV, 18-Mar-2025)

Ref Expression
Hypotheses pzriprng.r R = ring × 𝑠 ring
pzriprng.i I = × 0
Assertion pzriprnglem4 I SubGrp R

Proof

Step Hyp Ref Expression
1 pzriprng.r R = ring × 𝑠 ring
2 pzriprng.i I = × 0
3 0z 0
4 c0ex 0 V
5 4 snss 0 0
6 3 5 mpbi 0
7 xpss2 0 × 0 ×
8 6 7 ax-mp × 0 ×
9 1 pzriprnglem2 Base R = ×
10 8 2 9 3sstr4i I Base R
11 3 ne0ii
12 4 snnz 0
13 11 12 pm3.2i 0
14 xpnz 0 × 0
15 13 14 mpbi × 0
16 2 15 eqnetri I
17 1 2 pzriprnglem3 x I a x = a 0
18 1 2 pzriprnglem3 y I b y = b 0
19 simpr a x = a 0 x = a 0
20 19 adantr a x = a 0 b x = a 0
21 id y = b 0 y = b 0
22 20 21 oveqan12d a x = a 0 b y = b 0 x + R y = a 0 + R b 0
23 zringbas = Base ring
24 zringring ring Ring
25 24 a1i a b ring Ring
26 simpl a b a
27 3 a1i a b 0
28 simpr a b b
29 zaddcl a b a + b
30 00id 0 + 0 = 0
31 30 3 eqeltri 0 + 0
32 31 a1i a b 0 + 0
33 zringplusg + = + ring
34 eqid + R = + R
35 1 23 23 25 25 26 27 28 27 29 32 33 33 34 xpsadd a b a 0 + R b 0 = a + b 0 + 0
36 4 snid 0 0
37 30 36 eqeltri 0 + 0 0
38 2 eleq2i a + b 0 + 0 I a + b 0 + 0 × 0
39 opelxp a + b 0 + 0 × 0 a + b 0 + 0 0
40 38 39 bitri a + b 0 + 0 I a + b 0 + 0 0
41 29 37 40 sylanblrc a b a + b 0 + 0 I
42 35 41 eqeltrd a b a 0 + R b 0 I
43 42 ad4ant13 a x = a 0 b y = b 0 a 0 + R b 0 I
44 22 43 eqeltrd a x = a 0 b y = b 0 x + R y I
45 44 rexlimdva2 a x = a 0 b y = b 0 x + R y I
46 18 45 biimtrid a x = a 0 y I x + R y I
47 46 ralrimiv a x = a 0 y I x + R y I
48 zringgrp ring Grp
49 48 a1i a ring Grp
50 id a a
51 3 a1i a 0
52 eqid inv g ring = inv g ring
53 eqid inv g R = inv g R
54 1 23 23 49 49 50 51 52 52 53 xpsinv a inv g R a 0 = inv g ring a inv g ring 0
55 zringinvg a a = inv g ring a
56 znegcl a a
57 55 56 eqeltrrd a inv g ring a
58 neg0 0 = 0
59 58 36 eqeltri 0 0
60 zringinvg 0 0 = inv g ring 0
61 60 eleq1d 0 0 0 inv g ring 0 0
62 3 61 mp1i a 0 0 inv g ring 0 0
63 59 62 mpbii a inv g ring 0 0
64 57 63 opelxpd a inv g ring a inv g ring 0 × 0
65 54 64 eqeltrd a inv g R a 0 × 0
66 65 adantr a x = a 0 inv g R a 0 × 0
67 fveq2 x = a 0 inv g R x = inv g R a 0
68 67 adantl a x = a 0 inv g R x = inv g R a 0
69 2 a1i a x = a 0 I = × 0
70 66 68 69 3eltr4d a x = a 0 inv g R x I
71 47 70 jca a x = a 0 y I x + R y I inv g R x I
72 71 rexlimiva a x = a 0 y I x + R y I inv g R x I
73 17 72 sylbi x I y I x + R y I inv g R x I
74 73 rgen x I y I x + R y I inv g R x I
75 1 pzriprnglem1 R Rng
76 rnggrp R Rng R Grp
77 75 76 ax-mp R Grp
78 eqid Base R = Base R
79 78 34 53 issubg2 R Grp I SubGrp R I Base R I x I y I x + R y I inv g R x I
80 77 79 ax-mp I SubGrp R I Base R I x I y I x + R y I inv g R x I
81 10 16 74 80 mpbir3an I SubGrp R