Metamath Proof Explorer


Theorem ressply1mul

Description: A restricted polynomial algebra has the same multiplication operation. (Contributed by Mario Carneiro, 3-Jul-2015)

Ref Expression
Hypotheses ressply1.s S = Poly 1 R
ressply1.h H = R 𝑠 T
ressply1.u U = Poly 1 H
ressply1.b B = Base U
ressply1.2 φ T SubRing R
ressply1.p P = S 𝑠 B
Assertion ressply1mul φ X B Y B X U Y = X P Y

Proof

Step Hyp Ref Expression
1 ressply1.s S = Poly 1 R
2 ressply1.h H = R 𝑠 T
3 ressply1.u U = Poly 1 H
4 ressply1.b B = Base U
5 ressply1.2 φ T SubRing R
6 ressply1.p P = S 𝑠 B
7 eqid 1 𝑜 mPoly R = 1 𝑜 mPoly R
8 eqid 1 𝑜 mPoly H = 1 𝑜 mPoly H
9 3 4 ply1bas B = Base 1 𝑜 mPoly H
10 1on 1 𝑜 On
11 10 a1i φ 1 𝑜 On
12 eqid 1 𝑜 mPoly R 𝑠 B = 1 𝑜 mPoly R 𝑠 B
13 7 2 8 9 11 5 12 ressmplmul φ X B Y B X 1 𝑜 mPoly H Y = X 1 𝑜 mPoly R 𝑠 B Y
14 eqid U = U
15 3 8 14 ply1mulr U = 1 𝑜 mPoly H
16 15 oveqi X U Y = X 1 𝑜 mPoly H Y
17 eqid S = S
18 1 7 17 ply1mulr S = 1 𝑜 mPoly R
19 4 fvexi B V
20 6 17 ressmulr B V S = P
21 19 20 ax-mp S = P
22 eqid 1 𝑜 mPoly R = 1 𝑜 mPoly R
23 12 22 ressmulr B V 1 𝑜 mPoly R = 1 𝑜 mPoly R 𝑠 B
24 19 23 ax-mp 1 𝑜 mPoly R = 1 𝑜 mPoly R 𝑠 B
25 18 21 24 3eqtr3i P = 1 𝑜 mPoly R 𝑠 B
26 25 oveqi X P Y = X 1 𝑜 mPoly R 𝑠 B Y
27 13 16 26 3eqtr4g φ X B Y B X U Y = X P Y