Metamath Proof Explorer


Theorem ressply1invg

Description: An element of a restricted polynomial algebra has the same group inverse. (Contributed by Thierry Arnoux, 30-Jan-2025)

Ref Expression
Hypotheses ressply.1 S = Poly 1 R
ressply.2 H = R 𝑠 T
ressply.3 U = Poly 1 H
ressply.4 B = Base U
ressply.5 φ T SubRing R
ressply1.1 P = S 𝑠 B
ressply1invg.1 φ X B
Assertion ressply1invg φ inv g U X = inv g P X

Proof

Step Hyp Ref Expression
1 ressply.1 S = Poly 1 R
2 ressply.2 H = R 𝑠 T
3 ressply.3 U = Poly 1 H
4 ressply.4 B = Base U
5 ressply.5 φ T SubRing R
6 ressply1.1 P = S 𝑠 B
7 ressply1invg.1 φ X B
8 1 2 3 4 5 6 ressply1bas φ B = Base P
9 1 2 3 4 5 6 ressply1add φ y B X B y + U X = y + P X
10 9 anassrs φ y B X B y + U X = y + P X
11 7 10 mpidan φ y B y + U X = y + P X
12 eqid 0 S = 0 S
13 1 2 3 4 5 12 ressply10g φ 0 S = 0 U
14 1 2 3 4 subrgply1 T SubRing R B SubRing S
15 subrgrcl B SubRing S S Ring
16 ringmnd S Ring S Mnd
17 5 14 15 16 4syl φ S Mnd
18 subrgsubg B SubRing S B SubGrp S
19 12 subg0cl B SubGrp S 0 S B
20 5 14 18 19 4syl φ 0 S B
21 eqid PwSer 1 H = PwSer 1 H
22 eqid Base PwSer 1 H = Base PwSer 1 H
23 eqid Base S = Base S
24 1 2 3 4 5 21 22 23 ressply1bas2 φ B = Base PwSer 1 H Base S
25 inss2 Base PwSer 1 H Base S Base S
26 24 25 eqsstrdi φ B Base S
27 6 23 12 ress0g S Mnd 0 S B B Base S 0 S = 0 P
28 17 20 26 27 syl3anc φ 0 S = 0 P
29 13 28 eqtr3d φ 0 U = 0 P
30 29 adantr φ y B 0 U = 0 P
31 11 30 eqeq12d φ y B y + U X = 0 U y + P X = 0 P
32 8 31 riotaeqbidva φ ι y B | y + U X = 0 U = ι y Base P | y + P X = 0 P
33 eqid + U = + U
34 eqid 0 U = 0 U
35 eqid inv g U = inv g U
36 4 33 34 35 grpinvval X B inv g U X = ι y B | y + U X = 0 U
37 7 36 syl φ inv g U X = ι y B | y + U X = 0 U
38 7 8 eleqtrd φ X Base P
39 eqid Base P = Base P
40 eqid + P = + P
41 eqid 0 P = 0 P
42 eqid inv g P = inv g P
43 39 40 41 42 grpinvval X Base P inv g P X = ι y Base P | y + P X = 0 P
44 38 43 syl φ inv g P X = ι y Base P | y + P X = 0 P
45 32 37 44 3eqtr4d φ inv g U X = inv g P X