Metamath Proof Explorer


Theorem dgrle

Description: Given an explicit expression for a polynomial, the degree is at most the highest term in the sum. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses dgrle.1 φFPolyS
dgrle.2 φN0
dgrle.3 φk0NA
dgrle.4 φF=zk=0NAzk
Assertion dgrle φdegFN

Proof

Step Hyp Ref Expression
1 dgrle.1 φFPolyS
2 dgrle.2 φN0
3 dgrle.3 φk0NA
4 dgrle.4 φF=zk=0NAzk
5 1 2 3 4 coeeq2 φcoeffF=k0ifkNA0
6 5 ad2antrr φm0¬mNcoeffF=k0ifkNA0
7 6 fveq1d φm0¬mNcoeffFm=k0ifkNA0m
8 nfcv _km
9 nfv k¬mN
10 nffvmpt1 _kk0ifkNA0m
11 10 nfeq1 kk0ifkNA0m=0
12 9 11 nfim k¬mNk0ifkNA0m=0
13 breq1 k=mkNmN
14 13 notbid k=m¬kN¬mN
15 fveqeq2 k=mk0ifkNA0k=0k0ifkNA0m=0
16 14 15 imbi12d k=m¬kNk0ifkNA0k=0¬mNk0ifkNA0m=0
17 iffalse ¬kNifkNA0=0
18 17 fveq2d ¬kNIifkNA0=I0
19 0cn 0
20 fvi 0I0=0
21 19 20 ax-mp I0=0
22 18 21 eqtrdi ¬kNIifkNA0=0
23 eqid k0ifkNA0=k0ifkNA0
24 23 fvmpt2i k0k0ifkNA0k=IifkNA0
25 24 eqeq1d k0k0ifkNA0k=0IifkNA0=0
26 22 25 imbitrrid k0¬kNk0ifkNA0k=0
27 8 12 16 26 vtoclgaf m0¬mNk0ifkNA0m=0
28 27 imp m0¬mNk0ifkNA0m=0
29 28 adantll φm0¬mNk0ifkNA0m=0
30 7 29 eqtrd φm0¬mNcoeffFm=0
31 30 ex φm0¬mNcoeffFm=0
32 31 necon1ad φm0coeffFm0mN
33 32 ralrimiva φm0coeffFm0mN
34 eqid coeffF=coeffF
35 34 coef3 FPolyScoeffF:0
36 1 35 syl φcoeffF:0
37 plyco0 N0coeffF:0coeffFN+1=0m0coeffFm0mN
38 2 36 37 syl2anc φcoeffFN+1=0m0coeffFm0mN
39 33 38 mpbird φcoeffFN+1=0
40 eqid degF=degF
41 34 40 dgrlb FPolySN0coeffFN+1=0degFN
42 1 2 39 41 syl3anc φdegFN