Metamath Proof Explorer


Theorem bccolsum

Description: A column-sum rule for binomial coefficents. (Contributed by Scott Fenton, 24-Jun-2020)

Ref Expression
Assertion bccolsum N0C0k=0N(kC)=(N+1C+1)

Proof

Step Hyp Ref Expression
1 oveq2 m=00m=00
2 1 sumeq1d m=0k=0m(kC)=k=00(kC)
3 oveq1 m=0m+1=0+1
4 0p1e1 0+1=1
5 3 4 eqtrdi m=0m+1=1
6 5 oveq1d m=0(m+1C+1)=(1C+1)
7 2 6 eqeq12d m=0k=0m(kC)=(m+1C+1)k=00(kC)=(1C+1)
8 7 imbi2d m=0C0k=0m(kC)=(m+1C+1)C0k=00(kC)=(1C+1)
9 oveq2 m=n0m=0n
10 9 sumeq1d m=nk=0m(kC)=k=0n(kC)
11 oveq1 m=nm+1=n+1
12 11 oveq1d m=n(m+1C+1)=(n+1C+1)
13 10 12 eqeq12d m=nk=0m(kC)=(m+1C+1)k=0n(kC)=(n+1C+1)
14 13 imbi2d m=nC0k=0m(kC)=(m+1C+1)C0k=0n(kC)=(n+1C+1)
15 oveq2 m=n+10m=0n+1
16 15 sumeq1d m=n+1k=0m(kC)=k=0n+1(kC)
17 oveq1 m=n+1m+1=n+1+1
18 17 oveq1d m=n+1(m+1C+1)=(n+1+1C+1)
19 16 18 eqeq12d m=n+1k=0m(kC)=(m+1C+1)k=0n+1(kC)=(n+1+1C+1)
20 19 imbi2d m=n+1C0k=0m(kC)=(m+1C+1)C0k=0n+1(kC)=(n+1+1C+1)
21 oveq2 m=N0m=0N
22 21 sumeq1d m=Nk=0m(kC)=k=0N(kC)
23 oveq1 m=Nm+1=N+1
24 23 oveq1d m=N(m+1C+1)=(N+1C+1)
25 22 24 eqeq12d m=Nk=0m(kC)=(m+1C+1)k=0N(kC)=(N+1C+1)
26 25 imbi2d m=NC0k=0m(kC)=(m+1C+1)C0k=0N(kC)=(N+1C+1)
27 0z 0
28 0nn0 00
29 nn0z C0C
30 bccl 00C(0C)0
31 28 29 30 sylancr C0(0C)0
32 31 nn0cnd C0(0C)
33 oveq1 k=0(kC)=(0C)
34 33 fsum1 0(0C)k=00(kC)=(0C)
35 27 32 34 sylancr C0k=00(kC)=(0C)
36 elnn0 C0CC=0
37 1red C1
38 nnrp CC+
39 37 38 ltaddrp2d C1<C+1
40 peano2nn CC+1
41 40 nnred CC+1
42 37 41 ltnled C1<C+1¬C+11
43 39 42 mpbid C¬C+11
44 elfzle2 C+101C+11
45 43 44 nsyl C¬C+101
46 45 iffalsed CifC+1011!1C+1!C+1!0=0
47 1nn0 10
48 40 nnzd CC+1
49 bcval 10C+1(1C+1)=ifC+1011!1C+1!C+1!0
50 47 48 49 sylancr C(1C+1)=ifC+1011!1C+1!C+1!0
51 bc0k C(0C)=0
52 46 50 51 3eqtr4rd C(0C)=(1C+1)
53 bcnn 00(00)=1
54 28 53 ax-mp (00)=1
55 bcnn 10(11)=1
56 47 55 ax-mp (11)=1
57 54 56 eqtr4i (00)=(11)
58 oveq2 C=0(0C)=(00)
59 oveq1 C=0C+1=0+1
60 59 4 eqtrdi C=0C+1=1
61 60 oveq2d C=0(1C+1)=(11)
62 57 58 61 3eqtr4a C=0(0C)=(1C+1)
63 52 62 jaoi CC=0(0C)=(1C+1)
64 36 63 sylbi C0(0C)=(1C+1)
65 35 64 eqtrd C0k=00(kC)=(1C+1)
66 elnn0uz n0n0
67 66 biimpi n0n0
68 67 adantr n0C0n0
69 elfznn0 k0n+1k0
70 69 adantl n0C0k0n+1k0
71 simplr n0C0k0n+1C0
72 71 nn0zd n0C0k0n+1C
73 bccl k0C(kC)0
74 70 72 73 syl2anc n0C0k0n+1(kC)0
75 74 nn0cnd n0C0k0n+1(kC)
76 oveq1 k=n+1(kC)=(n+1C)
77 68 75 76 fsump1 n0C0k=0n+1(kC)=k=0n(kC)+(n+1C)
78 77 adantr n0C0k=0n(kC)=(n+1C+1)k=0n+1(kC)=k=0n(kC)+(n+1C)
79 id k=0n(kC)=(n+1C+1)k=0n(kC)=(n+1C+1)
80 nn0cn C0C
81 80 adantl n0C0C
82 1cnd n0C01
83 81 82 pncand n0C0C+1-1=C
84 83 oveq2d n0C0(n+1C+1-1)=(n+1C)
85 84 eqcomd n0C0(n+1C)=(n+1C+1-1)
86 79 85 oveqan12rd n0C0k=0n(kC)=(n+1C+1)k=0n(kC)+(n+1C)=(n+1C+1)+(n+1C+1-1)
87 peano2nn0 n0n+10
88 peano2nn0 C0C+10
89 88 nn0zd C0C+1
90 bcpasc n+10C+1(n+1C+1)+(n+1C+1-1)=(n+1+1C+1)
91 87 89 90 syl2an n0C0(n+1C+1)+(n+1C+1-1)=(n+1+1C+1)
92 91 adantr n0C0k=0n(kC)=(n+1C+1)(n+1C+1)+(n+1C+1-1)=(n+1+1C+1)
93 78 86 92 3eqtrd n0C0k=0n(kC)=(n+1C+1)k=0n+1(kC)=(n+1+1C+1)
94 93 exp31 n0C0k=0n(kC)=(n+1C+1)k=0n+1(kC)=(n+1+1C+1)
95 94 a2d n0C0k=0n(kC)=(n+1C+1)C0k=0n+1(kC)=(n+1+1C+1)
96 8 14 20 26 65 95 nn0ind N0C0k=0N(kC)=(N+1C+1)
97 96 imp N0C0k=0N(kC)=(N+1C+1)