Metamath Proof Explorer


Theorem coeval

Description: Value of the coefficient function. (Contributed by Mario Carneiro, 22-Jul-2014)

Ref Expression
Assertion coeval F Poly S coeff F = ι a 0 | n 0 a n + 1 = 0 F = z k = 0 n a k z k

Proof

Step Hyp Ref Expression
1 plyssc Poly S Poly
2 1 sseli F Poly S F Poly
3 eqeq1 f = F f = z k = 0 n a k z k F = z k = 0 n a k z k
4 3 anbi2d f = F a n + 1 = 0 f = z k = 0 n a k z k a n + 1 = 0 F = z k = 0 n a k z k
5 4 rexbidv f = F n 0 a n + 1 = 0 f = z k = 0 n a k z k n 0 a n + 1 = 0 F = z k = 0 n a k z k
6 5 riotabidv f = F ι a 0 | n 0 a n + 1 = 0 f = z k = 0 n a k z k = ι a 0 | n 0 a n + 1 = 0 F = z k = 0 n a k z k
7 df-coe coeff = f Poly ι a 0 | n 0 a n + 1 = 0 f = z k = 0 n a k z k
8 riotaex ι a 0 | n 0 a n + 1 = 0 F = z k = 0 n a k z k V
9 6 7 8 fvmpt F Poly coeff F = ι a 0 | n 0 a n + 1 = 0 F = z k = 0 n a k z k
10 2 9 syl F Poly S coeff F = ι a 0 | n 0 a n + 1 = 0 F = z k = 0 n a k z k