Metamath Proof Explorer


Theorem elplyr

Description: Sufficient condition for elementhood in the set of polynomials. (Contributed by Mario Carneiro, 17-Jul-2014) (Revised by Mario Carneiro, 23-Aug-2014)

Ref Expression
Assertion elplyr S N 0 A : 0 S z k = 0 N A k z k Poly S

Proof

Step Hyp Ref Expression
1 simp1 S N 0 A : 0 S S
2 simp2 S N 0 A : 0 S N 0
3 simp3 S N 0 A : 0 S A : 0 S
4 ssun1 S S 0
5 fss A : 0 S S S 0 A : 0 S 0
6 3 4 5 sylancl S N 0 A : 0 S A : 0 S 0
7 0cnd S N 0 A : 0 S 0
8 7 snssd S N 0 A : 0 S 0
9 1 8 unssd S N 0 A : 0 S S 0
10 cnex V
11 ssexg S 0 V S 0 V
12 9 10 11 sylancl S N 0 A : 0 S S 0 V
13 nn0ex 0 V
14 elmapg S 0 V 0 V A S 0 0 A : 0 S 0
15 12 13 14 sylancl S N 0 A : 0 S A S 0 0 A : 0 S 0
16 6 15 mpbird S N 0 A : 0 S A S 0 0
17 eqidd S N 0 A : 0 S z k = 0 N A k z k = z k = 0 N A k z k
18 oveq2 n = N 0 n = 0 N
19 18 sumeq1d n = N k = 0 n a k z k = k = 0 N a k z k
20 19 mpteq2dv n = N z k = 0 n a k z k = z k = 0 N a k z k
21 20 eqeq2d n = N z k = 0 N A k z k = z k = 0 n a k z k z k = 0 N A k z k = z k = 0 N a k z k
22 fveq1 a = A a k = A k
23 22 oveq1d a = A a k z k = A k z k
24 23 sumeq2sdv a = A k = 0 N a k z k = k = 0 N A k z k
25 24 mpteq2dv a = A z k = 0 N a k z k = z k = 0 N A k z k
26 25 eqeq2d a = A z k = 0 N A k z k = z k = 0 N a k z k z k = 0 N A k z k = z k = 0 N A k z k
27 21 26 rspc2ev N 0 A S 0 0 z k = 0 N A k z k = z k = 0 N A k z k n 0 a S 0 0 z k = 0 N A k z k = z k = 0 n a k z k
28 2 16 17 27 syl3anc S N 0 A : 0 S n 0 a S 0 0 z k = 0 N A k z k = z k = 0 n a k z k
29 elply z k = 0 N A k z k Poly S S n 0 a S 0 0 z k = 0 N A k z k = z k = 0 n a k z k
30 1 28 29 sylanbrc S N 0 A : 0 S z k = 0 N A k z k Poly S