Metamath Proof Explorer


Theorem evl1var

Description: Polynomial evaluation maps the variable to the identity function. (Contributed by Mario Carneiro, 12-Jun-2015)

Ref Expression
Hypotheses evl1var.q O = eval 1 R
evl1var.v X = var 1 R
evl1var.b B = Base R
Assertion evl1var R CRing O X = I B

Proof

Step Hyp Ref Expression
1 evl1var.q O = eval 1 R
2 evl1var.v X = var 1 R
3 evl1var.b B = Base R
4 crngring R CRing R Ring
5 eqid Poly 1 R = Poly 1 R
6 eqid Base Poly 1 R = Base Poly 1 R
7 2 5 6 vr1cl R Ring X Base Poly 1 R
8 4 7 syl R CRing X Base Poly 1 R
9 eqid 1 𝑜 eval R = 1 𝑜 eval R
10 eqid 1 𝑜 mPoly R = 1 𝑜 mPoly R
11 5 6 ply1bas Base Poly 1 R = Base 1 𝑜 mPoly R
12 1 9 3 10 11 evl1val R CRing X Base Poly 1 R O X = 1 𝑜 eval R X y B 1 𝑜 × y
13 8 12 mpdan R CRing O X = 1 𝑜 eval R X y B 1 𝑜 × y
14 df1o2 1 𝑜 =
15 3 fvexi B V
16 0ex V
17 eqid z B 1 𝑜 z = z B 1 𝑜 z
18 14 15 16 17 mapsncnv z B 1 𝑜 z -1 = y B 1 𝑜 × y
19 18 coeq2i 1 𝑜 eval R X z B 1 𝑜 z -1 = 1 𝑜 eval R X y B 1 𝑜 × y
20 3 ressid R CRing R 𝑠 B = R
21 20 oveq2d R CRing 1 𝑜 mVar R 𝑠 B = 1 𝑜 mVar R
22 21 fveq1d R CRing 1 𝑜 mVar R 𝑠 B = 1 𝑜 mVar R
23 2 vr1val X = 1 𝑜 mVar R
24 22 23 eqtr4di R CRing 1 𝑜 mVar R 𝑠 B = X
25 24 fveq2d R CRing 1 𝑜 eval R 1 𝑜 mVar R 𝑠 B = 1 𝑜 eval R X
26 9 3 evlval 1 𝑜 eval R = 1 𝑜 evalSub R B
27 eqid 1 𝑜 mVar R 𝑠 B = 1 𝑜 mVar R 𝑠 B
28 eqid R 𝑠 B = R 𝑠 B
29 1on 1 𝑜 On
30 29 a1i R CRing 1 𝑜 On
31 id R CRing R CRing
32 3 subrgid R Ring B SubRing R
33 4 32 syl R CRing B SubRing R
34 0lt1o 1 𝑜
35 34 a1i R CRing 1 𝑜
36 26 27 28 3 30 31 33 35 evlsvar R CRing 1 𝑜 eval R 1 𝑜 mVar R 𝑠 B = z B 1 𝑜 z
37 25 36 eqtr3d R CRing 1 𝑜 eval R X = z B 1 𝑜 z
38 37 coeq1d R CRing 1 𝑜 eval R X z B 1 𝑜 z -1 = z B 1 𝑜 z z B 1 𝑜 z -1
39 19 38 eqtr3id R CRing 1 𝑜 eval R X y B 1 𝑜 × y = z B 1 𝑜 z z B 1 𝑜 z -1
40 14 15 16 17 mapsnf1o2 z B 1 𝑜 z : B 1 𝑜 1-1 onto B
41 f1ococnv2 z B 1 𝑜 z : B 1 𝑜 1-1 onto B z B 1 𝑜 z z B 1 𝑜 z -1 = I B
42 40 41 mp1i R CRing z B 1 𝑜 z z B 1 𝑜 z -1 = I B
43 13 39 42 3eqtrd R CRing O X = I B