Metamath Proof Explorer


Theorem plyval

Description: Value of the polynomial set function. (Contributed by Mario Carneiro, 17-Jul-2014)

Ref Expression
Assertion plyval S Poly S = f | n 0 a S 0 0 f = z k = 0 n a k z k

Proof

Step Hyp Ref Expression
1 cnex V
2 1 elpw2 S 𝒫 S
3 uneq1 x = S x 0 = S 0
4 3 oveq1d x = S x 0 0 = S 0 0
5 4 rexeqdv x = S a x 0 0 f = z k = 0 n a k z k a S 0 0 f = z k = 0 n a k z k
6 5 rexbidv x = S n 0 a x 0 0 f = z k = 0 n a k z k n 0 a S 0 0 f = z k = 0 n a k z k
7 6 abbidv x = S f | n 0 a x 0 0 f = z k = 0 n a k z k = f | n 0 a S 0 0 f = z k = 0 n a k z k
8 df-ply Poly = x 𝒫 f | n 0 a x 0 0 f = z k = 0 n a k z k
9 nn0ex 0 V
10 ovex S 0 0 V
11 9 10 ab2rexex f | n 0 a S 0 0 f = z k = 0 n a k z k V
12 7 8 11 fvmpt S 𝒫 Poly S = f | n 0 a S 0 0 f = z k = 0 n a k z k
13 2 12 sylbir S Poly S = f | n 0 a S 0 0 f = z k = 0 n a k z k