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