Metamath Proof Explorer


Theorem deg1val

Description: Value of the univariate degree as a supremum. (Contributed by Stefan O'Rear, 29-Mar-2015) (Revised by AV, 25-Jul-2019)

Ref Expression
Hypotheses deg1leb.d D = deg 1 R
deg1leb.p P = Poly 1 R
deg1leb.b B = Base P
deg1leb.y 0 ˙ = 0 R
deg1leb.a A = coe 1 F
Assertion deg1val F B D F = sup A supp 0 ˙ * <

Proof

Step Hyp Ref Expression
1 deg1leb.d D = deg 1 R
2 deg1leb.p P = Poly 1 R
3 deg1leb.b B = Base P
4 deg1leb.y 0 ˙ = 0 R
5 deg1leb.a A = coe 1 F
6 1 deg1fval D = 1 𝑜 mDeg R
7 eqid 1 𝑜 mPoly R = 1 𝑜 mPoly R
8 eqid PwSer 1 R = PwSer 1 R
9 2 8 3 ply1bas B = Base 1 𝑜 mPoly R
10 psr1baslem 0 1 𝑜 = y 0 1 𝑜 | y -1 Fin
11 tdeglem2 x 0 1 𝑜 x = x 0 1 𝑜 fld x
12 6 7 9 4 10 11 mdegval F B D F = sup x 0 1 𝑜 x F supp 0 ˙ * <
13 4 fvexi 0 ˙ V
14 suppimacnv F B 0 ˙ V F supp 0 ˙ = F -1 V 0 ˙
15 13 14 mpan2 F B F supp 0 ˙ = F -1 V 0 ˙
16 15 imaeq2d F B x 0 1 𝑜 x F supp 0 ˙ = x 0 1 𝑜 x F -1 V 0 ˙
17 imaco x 0 1 𝑜 x F -1 V 0 ˙ = x 0 1 𝑜 x F -1 V 0 ˙
18 16 17 eqtr4di F B x 0 1 𝑜 x F supp 0 ˙ = x 0 1 𝑜 x F -1 V 0 ˙
19 df1o2 1 𝑜 =
20 nn0ex 0 V
21 0ex V
22 eqid x 0 1 𝑜 x = x 0 1 𝑜 x
23 19 20 21 22 mapsncnv x 0 1 𝑜 x -1 = y 0 1 𝑜 × y
24 5 3 2 23 coe1fval2 F B A = F x 0 1 𝑜 x -1
25 24 cnveqd F B A -1 = F x 0 1 𝑜 x -1 -1
26 cnvco F x 0 1 𝑜 x -1 -1 = x 0 1 𝑜 x -1 -1 F -1
27 cocnvcnv1 x 0 1 𝑜 x -1 -1 F -1 = x 0 1 𝑜 x F -1
28 26 27 eqtri F x 0 1 𝑜 x -1 -1 = x 0 1 𝑜 x F -1
29 25 28 eqtr2di F B x 0 1 𝑜 x F -1 = A -1
30 29 imaeq1d F B x 0 1 𝑜 x F -1 V 0 ˙ = A -1 V 0 ˙
31 18 30 eqtrd F B x 0 1 𝑜 x F supp 0 ˙ = A -1 V 0 ˙
32 5 fvexi A V
33 suppimacnv A V 0 ˙ V A supp 0 ˙ = A -1 V 0 ˙
34 33 eqcomd A V 0 ˙ V A -1 V 0 ˙ = A supp 0 ˙
35 32 13 34 mp2an A -1 V 0 ˙ = A supp 0 ˙
36 31 35 eqtrdi F B x 0 1 𝑜 x F supp 0 ˙ = A supp 0 ˙
37 36 supeq1d F B sup x 0 1 𝑜 x F supp 0 ˙ * < = sup A supp 0 ˙ * <
38 12 37 eqtrd F B D F = sup A supp 0 ˙ * <