Metamath Proof Explorer


Theorem elplyd

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

Ref Expression
Hypotheses elplyd.1 φ S
elplyd.2 φ N 0
elplyd.3 φ k 0 N A S
Assertion elplyd φ z k = 0 N A z k Poly S

Proof

Step Hyp Ref Expression
1 elplyd.1 φ S
2 elplyd.2 φ N 0
3 elplyd.3 φ k 0 N A S
4 nffvmpt1 _ k k 0 if k 0 N A 0 j
5 nfcv _ k ×
6 nfcv _ k z j
7 4 5 6 nfov _ k k 0 if k 0 N A 0 j z j
8 nfcv _ j k 0 if k 0 N A 0 k z k
9 fveq2 j = k k 0 if k 0 N A 0 j = k 0 if k 0 N A 0 k
10 oveq2 j = k z j = z k
11 9 10 oveq12d j = k k 0 if k 0 N A 0 j z j = k 0 if k 0 N A 0 k z k
12 7 8 11 cbvsumi j = 0 N k 0 if k 0 N A 0 j z j = k = 0 N k 0 if k 0 N A 0 k z k
13 elfznn0 k 0 N k 0
14 iftrue k 0 N if k 0 N A 0 = A
15 14 adantl φ k 0 N if k 0 N A 0 = A
16 15 3 eqeltrd φ k 0 N if k 0 N A 0 S
17 eqid k 0 if k 0 N A 0 = k 0 if k 0 N A 0
18 17 fvmpt2 k 0 if k 0 N A 0 S k 0 if k 0 N A 0 k = if k 0 N A 0
19 13 16 18 syl2an2 φ k 0 N k 0 if k 0 N A 0 k = if k 0 N A 0
20 19 15 eqtrd φ k 0 N k 0 if k 0 N A 0 k = A
21 20 oveq1d φ k 0 N k 0 if k 0 N A 0 k z k = A z k
22 21 sumeq2dv φ k = 0 N k 0 if k 0 N A 0 k z k = k = 0 N A z k
23 12 22 eqtrid φ j = 0 N k 0 if k 0 N A 0 j z j = k = 0 N A z k
24 23 mpteq2dv φ z j = 0 N k 0 if k 0 N A 0 j z j = z k = 0 N A z k
25 0cnd φ 0
26 25 snssd φ 0
27 1 26 unssd φ S 0
28 elun1 A S A S 0
29 3 28 syl φ k 0 N A S 0
30 29 adantlr φ k 0 k 0 N A S 0
31 ssun2 0 S 0
32 c0ex 0 V
33 32 snss 0 S 0 0 S 0
34 31 33 mpbir 0 S 0
35 34 a1i φ k 0 ¬ k 0 N 0 S 0
36 30 35 ifclda φ k 0 if k 0 N A 0 S 0
37 36 fmpttd φ k 0 if k 0 N A 0 : 0 S 0
38 elplyr S 0 N 0 k 0 if k 0 N A 0 : 0 S 0 z j = 0 N k 0 if k 0 N A 0 j z j Poly S 0
39 27 2 37 38 syl3anc φ z j = 0 N k 0 if k 0 N A 0 j z j Poly S 0
40 24 39 eqeltrrd φ z k = 0 N A z k Poly S 0
41 plyun0 Poly S 0 = Poly S
42 40 41 eleqtrdi φ z k = 0 N A z k Poly S