Metamath Proof Explorer


Theorem prjcrvfval

Description: Value of the projective curve function. (Contributed by SN, 23-Nov-2024)

Ref Expression
Hypotheses prjcrvfval.h H = 0 N mHomP K
prjcrvfval.e E = 0 N eval K
prjcrvfval.p P = N ℙ𝕣𝕠𝕛 n K
prjcrvfval.0 0 ˙ = 0 K
prjcrvfval.n φ N 0
prjcrvfval.k φ K Field
Assertion prjcrvfval φ N PrjCrv K = f ran H p P | E f p = 0 ˙

Proof

Step Hyp Ref Expression
1 prjcrvfval.h H = 0 N mHomP K
2 prjcrvfval.e E = 0 N eval K
3 prjcrvfval.p P = N ℙ𝕣𝕠𝕛 n K
4 prjcrvfval.0 0 ˙ = 0 K
5 prjcrvfval.n φ N 0
6 prjcrvfval.k φ K Field
7 oveq2 n = N 0 n = 0 N
8 oveq12 0 n = 0 N k = K 0 n mHomP k = 0 N mHomP K
9 7 8 sylan n = N k = K 0 n mHomP k = 0 N mHomP K
10 9 1 eqtr4di n = N k = K 0 n mHomP k = H
11 10 rneqd n = N k = K ran 0 n mHomP k = ran H
12 11 unieqd n = N k = K ran 0 n mHomP k = ran H
13 oveq12 n = N k = K n ℙ𝕣𝕠𝕛 n k = N ℙ𝕣𝕠𝕛 n K
14 13 3 eqtr4di n = N k = K n ℙ𝕣𝕠𝕛 n k = P
15 id k = K k = K
16 7 15 oveqan12d n = N k = K 0 n eval k = 0 N eval K
17 16 2 eqtr4di n = N k = K 0 n eval k = E
18 17 fveq1d n = N k = K 0 n eval k f = E f
19 18 imaeq1d n = N k = K 0 n eval k f p = E f p
20 fveq2 k = K 0 k = 0 K
21 20 4 eqtr4di k = K 0 k = 0 ˙
22 21 adantl n = N k = K 0 k = 0 ˙
23 22 sneqd n = N k = K 0 k = 0 ˙
24 19 23 eqeq12d n = N k = K 0 n eval k f p = 0 k E f p = 0 ˙
25 14 24 rabeqbidv n = N k = K p n ℙ𝕣𝕠𝕛 n k | 0 n eval k f p = 0 k = p P | E f p = 0 ˙
26 12 25 mpteq12dv n = N k = K f ran 0 n mHomP k p n ℙ𝕣𝕠𝕛 n k | 0 n eval k f p = 0 k = f ran H p P | E f p = 0 ˙
27 df-prjcrv PrjCrv = n 0 , k Field f ran 0 n mHomP k p n ℙ𝕣𝕠𝕛 n k | 0 n eval k f p = 0 k
28 1 ovexi H V
29 28 rnex ran H V
30 29 uniex ran H V
31 30 mptex f ran H p P | E f p = 0 ˙ V
32 26 27 31 ovmpoa N 0 K Field N PrjCrv K = f ran H p P | E f p = 0 ˙
33 5 6 32 syl2anc φ N PrjCrv K = f ran H p P | E f p = 0 ˙