Metamath Proof Explorer


Theorem coeeq

Description: If A satisfies the properties of the coefficient function, it must be equal to the coefficient function. (Contributed by Mario Carneiro, 22-Jul-2014) (Revised by Mario Carneiro, 23-Aug-2014)

Ref Expression
Hypotheses coeeq.1 φFPolyS
coeeq.2 φN0
coeeq.3 φA:0
coeeq.4 φAN+1=0
coeeq.5 φF=zk=0NAkzk
Assertion coeeq φcoeffF=A

Proof

Step Hyp Ref Expression
1 coeeq.1 φFPolyS
2 coeeq.2 φN0
3 coeeq.3 φA:0
4 coeeq.4 φAN+1=0
5 coeeq.5 φF=zk=0NAkzk
6 coeval FPolyScoeffF=ιa0|n0an+1=0F=zk=0nakzk
7 1 6 syl φcoeffF=ιa0|n0an+1=0F=zk=0nakzk
8 fvoveq1 n=Nn+1=N+1
9 8 imaeq2d n=NAn+1=AN+1
10 9 eqeq1d n=NAn+1=0AN+1=0
11 oveq2 n=N0n=0N
12 11 sumeq1d n=Nk=0nAkzk=k=0NAkzk
13 12 mpteq2dv n=Nzk=0nAkzk=zk=0NAkzk
14 13 eqeq2d n=NF=zk=0nAkzkF=zk=0NAkzk
15 10 14 anbi12d n=NAn+1=0F=zk=0nAkzkAN+1=0F=zk=0NAkzk
16 15 rspcev N0AN+1=0F=zk=0NAkzkn0An+1=0F=zk=0nAkzk
17 2 4 5 16 syl12anc φn0An+1=0F=zk=0nAkzk
18 cnex V
19 nn0ex 0V
20 18 19 elmap A0A:0
21 3 20 sylibr φA0
22 coeeu FPolyS∃!a0n0an+1=0F=zk=0nakzk
23 1 22 syl φ∃!a0n0an+1=0F=zk=0nakzk
24 imaeq1 a=Aan+1=An+1
25 24 eqeq1d a=Aan+1=0An+1=0
26 fveq1 a=Aak=Ak
27 26 oveq1d a=Aakzk=Akzk
28 27 sumeq2sdv a=Ak=0nakzk=k=0nAkzk
29 28 mpteq2dv a=Azk=0nakzk=zk=0nAkzk
30 29 eqeq2d a=AF=zk=0nakzkF=zk=0nAkzk
31 25 30 anbi12d a=Aan+1=0F=zk=0nakzkAn+1=0F=zk=0nAkzk
32 31 rexbidv a=An0an+1=0F=zk=0nakzkn0An+1=0F=zk=0nAkzk
33 32 riota2 A0∃!a0n0an+1=0F=zk=0nakzkn0An+1=0F=zk=0nAkzkιa0|n0an+1=0F=zk=0nakzk=A
34 21 23 33 syl2anc φn0An+1=0F=zk=0nAkzkιa0|n0an+1=0F=zk=0nakzk=A
35 17 34 mpbid φιa0|n0an+1=0F=zk=0nakzk=A
36 7 35 eqtrd φcoeffF=A