Metamath Proof Explorer


Theorem plyaddlem

Description: Lemma for plyadd . (Contributed by Mario Carneiro, 21-Jul-2014)

Ref Expression
Hypotheses plyadd.1 φ F Poly S
plyadd.2 φ G Poly S
plyadd.3 φ x S y S x + y S
plyadd.m φ M 0
plyadd.n φ N 0
plyadd.a φ A S 0 0
plyadd.b φ B S 0 0
plyadd.a2 φ A M + 1 = 0
plyadd.b2 φ B N + 1 = 0
plyadd.f φ F = z k = 0 M A k z k
plyadd.g φ G = z k = 0 N B k z k
Assertion plyaddlem φ F + f G Poly S

Proof

Step Hyp Ref Expression
1 plyadd.1 φ F Poly S
2 plyadd.2 φ G Poly S
3 plyadd.3 φ x S y S x + y S
4 plyadd.m φ M 0
5 plyadd.n φ N 0
6 plyadd.a φ A S 0 0
7 plyadd.b φ B S 0 0
8 plyadd.a2 φ A M + 1 = 0
9 plyadd.b2 φ B N + 1 = 0
10 plyadd.f φ F = z k = 0 M A k z k
11 plyadd.g φ G = z k = 0 N B k z k
12 plybss F Poly S S
13 1 12 syl φ S
14 0cnd φ 0
15 14 snssd φ 0
16 13 15 unssd φ S 0
17 cnex V
18 ssexg S 0 V S 0 V
19 16 17 18 sylancl φ S 0 V
20 nn0ex 0 V
21 elmapg S 0 V 0 V A S 0 0 A : 0 S 0
22 19 20 21 sylancl φ A S 0 0 A : 0 S 0
23 6 22 mpbid φ A : 0 S 0
24 23 16 fssd φ A : 0
25 elmapg S 0 V 0 V B S 0 0 B : 0 S 0
26 19 20 25 sylancl φ B S 0 0 B : 0 S 0
27 7 26 mpbid φ B : 0 S 0
28 27 16 fssd φ B : 0
29 1 2 4 5 24 28 8 9 10 11 plyaddlem1 φ F + f G = z k = 0 if M N N M A + f B k z k
30 5 4 ifcld φ if M N N M 0
31 eqid S 0 = S 0
32 13 31 3 un0addcl φ x S 0 y S 0 x + y S 0
33 20 a1i φ 0 V
34 inidm 0 0 = 0
35 32 23 27 33 33 34 off φ A + f B : 0 S 0
36 elfznn0 k 0 if M N N M k 0
37 ffvelrn A + f B : 0 S 0 k 0 A + f B k S 0
38 35 36 37 syl2an φ k 0 if M N N M A + f B k S 0
39 16 30 38 elplyd φ z k = 0 if M N N M A + f B k z k Poly S 0
40 29 39 eqeltrd φ F + f G Poly S 0
41 plyun0 Poly S 0 = Poly S
42 40 41 eleqtrdi φ F + f G Poly S