Metamath Proof Explorer


Theorem binom

Description: The binomial theorem: ( A + B ) ^ N is the sum from k = 0 to N of ( N _C k ) x. ( ( A ^ k ) x. ( B ^ ( N - k ) ) . Theorem 15-2.8 of Gleason p. 296. This part of the proof sets up the induction and does the base case, with the bulk of the work (the induction step) in binomlem . This is Metamath 100 proof #44. (Contributed by NM, 7-Dec-2005) (Proof shortened by Mario Carneiro, 24-Apr-2014)

Ref Expression
Assertion binom ABN0A+BN=k=0N(Nk)ANkBk

Proof

Step Hyp Ref Expression
1 oveq2 x=0A+Bx=A+B0
2 oveq2 x=00x=00
3 oveq1 x=0(xk)=(0k)
4 oveq1 x=0xk=0k
5 4 oveq2d x=0Axk=A0k
6 5 oveq1d x=0AxkBk=A0kBk
7 3 6 oveq12d x=0(xk)AxkBk=(0k)A0kBk
8 7 adantr x=0k0x(xk)AxkBk=(0k)A0kBk
9 2 8 sumeq12dv x=0k=0x(xk)AxkBk=k=00(0k)A0kBk
10 1 9 eqeq12d x=0A+Bx=k=0x(xk)AxkBkA+B0=k=00(0k)A0kBk
11 10 imbi2d x=0ABA+Bx=k=0x(xk)AxkBkABA+B0=k=00(0k)A0kBk
12 oveq2 x=nA+Bx=A+Bn
13 oveq2 x=n0x=0n
14 oveq1 x=n(xk)=(nk)
15 oveq1 x=nxk=nk
16 15 oveq2d x=nAxk=Ank
17 16 oveq1d x=nAxkBk=AnkBk
18 14 17 oveq12d x=n(xk)AxkBk=(nk)AnkBk
19 18 adantr x=nk0x(xk)AxkBk=(nk)AnkBk
20 13 19 sumeq12dv x=nk=0x(xk)AxkBk=k=0n(nk)AnkBk
21 12 20 eqeq12d x=nA+Bx=k=0x(xk)AxkBkA+Bn=k=0n(nk)AnkBk
22 21 imbi2d x=nABA+Bx=k=0x(xk)AxkBkABA+Bn=k=0n(nk)AnkBk
23 oveq2 x=n+1A+Bx=A+Bn+1
24 oveq2 x=n+10x=0n+1
25 oveq1 x=n+1(xk)=(n+1k)
26 oveq1 x=n+1xk=n+1-k
27 26 oveq2d x=n+1Axk=An+1-k
28 27 oveq1d x=n+1AxkBk=An+1-kBk
29 25 28 oveq12d x=n+1(xk)AxkBk=(n+1k)An+1-kBk
30 29 adantr x=n+1k0x(xk)AxkBk=(n+1k)An+1-kBk
31 24 30 sumeq12dv x=n+1k=0x(xk)AxkBk=k=0n+1(n+1k)An+1-kBk
32 23 31 eqeq12d x=n+1A+Bx=k=0x(xk)AxkBkA+Bn+1=k=0n+1(n+1k)An+1-kBk
33 32 imbi2d x=n+1ABA+Bx=k=0x(xk)AxkBkABA+Bn+1=k=0n+1(n+1k)An+1-kBk
34 oveq2 x=NA+Bx=A+BN
35 oveq2 x=N0x=0N
36 oveq1 x=N(xk)=(Nk)
37 oveq1 x=Nxk=Nk
38 37 oveq2d x=NAxk=ANk
39 38 oveq1d x=NAxkBk=ANkBk
40 36 39 oveq12d x=N(xk)AxkBk=(Nk)ANkBk
41 40 adantr x=Nk0x(xk)AxkBk=(Nk)ANkBk
42 35 41 sumeq12dv x=Nk=0x(xk)AxkBk=k=0N(Nk)ANkBk
43 34 42 eqeq12d x=NA+Bx=k=0x(xk)AxkBkA+BN=k=0N(Nk)ANkBk
44 43 imbi2d x=NABA+Bx=k=0x(xk)AxkBkABA+BN=k=0N(Nk)ANkBk
45 exp0 AA0=1
46 exp0 BB0=1
47 45 46 oveqan12d ABA0B0=11
48 1t1e1 11=1
49 47 48 eqtrdi ABA0B0=1
50 49 oveq2d AB1A0B0=11
51 50 48 eqtrdi AB1A0B0=1
52 0z 0
53 ax-1cn 1
54 51 53 eqeltrdi AB1A0B0
55 oveq2 k=0(0k)=(00)
56 0nn0 00
57 bcn0 00(00)=1
58 56 57 ax-mp (00)=1
59 55 58 eqtrdi k=0(0k)=1
60 oveq2 k=00k=00
61 0m0e0 00=0
62 60 61 eqtrdi k=00k=0
63 62 oveq2d k=0A0k=A0
64 oveq2 k=0Bk=B0
65 63 64 oveq12d k=0A0kBk=A0B0
66 59 65 oveq12d k=0(0k)A0kBk=1A0B0
67 66 fsum1 01A0B0k=00(0k)A0kBk=1A0B0
68 52 54 67 sylancr ABk=00(0k)A0kBk=1A0B0
69 addcl ABA+B
70 69 exp0d ABA+B0=1
71 51 68 70 3eqtr4rd ABA+B0=k=00(0k)A0kBk
72 simprl n0ABA
73 simprr n0ABB
74 simpl n0ABn0
75 id A+Bn=k=0n(nk)AnkBkA+Bn=k=0n(nk)AnkBk
76 72 73 74 75 binomlem n0ABA+Bn=k=0n(nk)AnkBkA+Bn+1=k=0n+1(n+1k)An+1-kBk
77 76 exp31 n0ABA+Bn=k=0n(nk)AnkBkA+Bn+1=k=0n+1(n+1k)An+1-kBk
78 77 a2d n0ABA+Bn=k=0n(nk)AnkBkABA+Bn+1=k=0n+1(n+1k)An+1-kBk
79 11 22 33 44 71 78 nn0ind N0ABA+BN=k=0N(Nk)ANkBk
80 79 impcom ABN0A+BN=k=0N(Nk)ANkBk
81 80 3impa ABN0A+BN=k=0N(Nk)ANkBk