Metamath Proof Explorer


Theorem ressply1sub

Description: A restricted polynomial algebra has the same subtraction operation. (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
ressply1sub.1 φ X B
ressply1sub.2 φ Y B
Assertion ressply1sub φ X - U Y = X - P Y

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 ressply1sub.1 φ X B
8 ressply1sub.2 φ Y B
9 1 2 3 4 5 6 8 ressply1invg φ inv g U Y = inv g P Y
10 9 oveq2d φ X + U inv g U Y = X + U inv g P Y
11 1 2 3 4 subrgply1 T SubRing R B SubRing S
12 subrgsubg B SubRing S B SubGrp S
13 6 subggrp B SubGrp S P Grp
14 5 11 12 13 4syl φ P Grp
15 1 2 3 4 5 6 ressply1bas φ B = Base P
16 8 15 eleqtrd φ Y Base P
17 eqid Base P = Base P
18 eqid inv g P = inv g P
19 17 18 grpinvcl P Grp Y Base P inv g P Y Base P
20 14 16 19 syl2anc φ inv g P Y Base P
21 20 15 eleqtrrd φ inv g P Y B
22 7 21 jca φ X B inv g P Y B
23 1 2 3 4 5 6 ressply1add φ X B inv g P Y B X + U inv g P Y = X + P inv g P Y
24 22 23 mpdan φ X + U inv g P Y = X + P inv g P Y
25 10 24 eqtrd φ X + U inv g U Y = X + P inv g P Y
26 eqid + U = + U
27 eqid inv g U = inv g U
28 eqid - U = - U
29 4 26 27 28 grpsubval X B Y B X - U Y = X + U inv g U Y
30 7 8 29 syl2anc φ X - U Y = X + U inv g U Y
31 7 15 eleqtrd φ X Base P
32 eqid + P = + P
33 eqid - P = - P
34 17 32 18 33 grpsubval X Base P Y Base P X - P Y = X + P inv g P Y
35 31 16 34 syl2anc φ X - P Y = X + P inv g P Y
36 25 30 35 3eqtr4d φ X - U Y = X - P Y