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Properties of real and complex numbers
bccolsum
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Theorem
bccolsum
Description:
A column-sum rule for binomial coefficents.
(Contributed by
Scott Fenton
, 24-Jun-2020)
Ref
Expression
Assertion
bccolsum
⊢
N
∈
ℕ
0
∧
C
∈
ℕ
0
→
∑
k
=
0
N
(
k
C
)
=
(
N
+
1
C
+
1
)
Proof
Step
Hyp
Ref
Expression
1
oveq2
⊢
m
=
0
→
0
…
m
=
0
…
0
2
1
sumeq1d
⊢
m
=
0
→
∑
k
=
0
m
(
k
C
)
=
∑
k
=
0
0
(
k
C
)
3
oveq1
⊢
m
=
0
→
m
+
1
=
0
+
1
4
0p1e1
⊢
0
+
1
=
1
5
3
4
eqtrdi
⊢
m
=
0
→
m
+
1
=
1
6
5
oveq1d
⊢
m
=
0
→
(
m
+
1
C
+
1
)
=
(
1
C
+
1
)
7
2
6
eqeq12d
⊢
m
=
0
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
∑
k
=
0
0
(
k
C
)
=
(
1
C
+
1
)
8
7
imbi2d
⊢
m
=
0
→
C
∈
ℕ
0
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
C
∈
ℕ
0
→
∑
k
=
0
0
(
k
C
)
=
(
1
C
+
1
)
9
oveq2
⊢
m
=
n
→
0
…
m
=
0
…
n
10
9
sumeq1d
⊢
m
=
n
→
∑
k
=
0
m
(
k
C
)
=
∑
k
=
0
n
(
k
C
)
11
oveq1
⊢
m
=
n
→
m
+
1
=
n
+
1
12
11
oveq1d
⊢
m
=
n
→
(
m
+
1
C
+
1
)
=
(
n
+
1
C
+
1
)
13
10
12
eqeq12d
⊢
m
=
n
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
14
13
imbi2d
⊢
m
=
n
→
C
∈
ℕ
0
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
C
∈
ℕ
0
→
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
15
oveq2
⊢
m
=
n
+
1
→
0
…
m
=
0
…
n
+
1
16
15
sumeq1d
⊢
m
=
n
+
1
→
∑
k
=
0
m
(
k
C
)
=
∑
k
=
0
n
+
1
(
k
C
)
17
oveq1
⊢
m
=
n
+
1
→
m
+
1
=
n
+
1
+
1
18
17
oveq1d
⊢
m
=
n
+
1
→
(
m
+
1
C
+
1
)
=
(
n
+
1
+
1
C
+
1
)
19
16
18
eqeq12d
⊢
m
=
n
+
1
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
∑
k
=
0
n
+
1
(
k
C
)
=
(
n
+
1
+
1
C
+
1
)
20
19
imbi2d
⊢
m
=
n
+
1
→
C
∈
ℕ
0
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
C
∈
ℕ
0
→
∑
k
=
0
n
+
1
(
k
C
)
=
(
n
+
1
+
1
C
+
1
)
21
oveq2
⊢
m
=
N
→
0
…
m
=
0
…
N
22
21
sumeq1d
⊢
m
=
N
→
∑
k
=
0
m
(
k
C
)
=
∑
k
=
0
N
(
k
C
)
23
oveq1
⊢
m
=
N
→
m
+
1
=
N
+
1
24
23
oveq1d
⊢
m
=
N
→
(
m
+
1
C
+
1
)
=
(
N
+
1
C
+
1
)
25
22
24
eqeq12d
⊢
m
=
N
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
∑
k
=
0
N
(
k
C
)
=
(
N
+
1
C
+
1
)
26
25
imbi2d
⊢
m
=
N
→
C
∈
ℕ
0
→
∑
k
=
0
m
(
k
C
)
=
(
m
+
1
C
+
1
)
↔
C
∈
ℕ
0
→
∑
k
=
0
N
(
k
C
)
=
(
N
+
1
C
+
1
)
27
0z
⊢
0
∈
ℤ
28
0nn0
⊢
0
∈
ℕ
0
29
nn0z
⊢
C
∈
ℕ
0
→
C
∈
ℤ
30
bccl
⊢
0
∈
ℕ
0
∧
C
∈
ℤ
→
(
0
C
)
∈
ℕ
0
31
28
29
30
sylancr
⊢
C
∈
ℕ
0
→
(
0
C
)
∈
ℕ
0
32
31
nn0cnd
⊢
C
∈
ℕ
0
→
(
0
C
)
∈
ℂ
33
oveq1
⊢
k
=
0
→
(
k
C
)
=
(
0
C
)
34
33
fsum1
⊢
0
∈
ℤ
∧
(
0
C
)
∈
ℂ
→
∑
k
=
0
0
(
k
C
)
=
(
0
C
)
35
27
32
34
sylancr
⊢
C
∈
ℕ
0
→
∑
k
=
0
0
(
k
C
)
=
(
0
C
)
36
elnn0
⊢
C
∈
ℕ
0
↔
C
∈
ℕ
∨
C
=
0
37
1red
⊢
C
∈
ℕ
→
1
∈
ℝ
38
nnrp
⊢
C
∈
ℕ
→
C
∈
ℝ
+
39
37
38
ltaddrp2d
⊢
C
∈
ℕ
→
1
<
C
+
1
40
peano2nn
⊢
C
∈
ℕ
→
C
+
1
∈
ℕ
41
40
nnred
⊢
C
∈
ℕ
→
C
+
1
∈
ℝ
42
37
41
ltnled
⊢
C
∈
ℕ
→
1
<
C
+
1
↔
¬
C
+
1
≤
1
43
39
42
mpbid
⊢
C
∈
ℕ
→
¬
C
+
1
≤
1
44
elfzle2
⊢
C
+
1
∈
0
…
1
→
C
+
1
≤
1
45
43
44
nsyl
⊢
C
∈
ℕ
→
¬
C
+
1
∈
0
…
1
46
45
iffalsed
⊢
C
∈
ℕ
→
if
C
+
1
∈
0
…
1
1
!
1
−
C
+
1
!
C
+
1
!
0
=
0
47
1nn0
⊢
1
∈
ℕ
0
48
40
nnzd
⊢
C
∈
ℕ
→
C
+
1
∈
ℤ
49
bcval
⊢
1
∈
ℕ
0
∧
C
+
1
∈
ℤ
→
(
1
C
+
1
)
=
if
C
+
1
∈
0
…
1
1
!
1
−
C
+
1
!
C
+
1
!
0
50
47
48
49
sylancr
⊢
C
∈
ℕ
→
(
1
C
+
1
)
=
if
C
+
1
∈
0
…
1
1
!
1
−
C
+
1
!
C
+
1
!
0
51
bc0k
⊢
C
∈
ℕ
→
(
0
C
)
=
0
52
46
50
51
3eqtr4rd
⊢
C
∈
ℕ
→
(
0
C
)
=
(
1
C
+
1
)
53
bcnn
⊢
0
∈
ℕ
0
→
(
0
0
)
=
1
54
28
53
ax-mp
⊢
(
0
0
)
=
1
55
bcnn
⊢
1
∈
ℕ
0
→
(
1
1
)
=
1
56
47
55
ax-mp
⊢
(
1
1
)
=
1
57
54
56
eqtr4i
⊢
(
0
0
)
=
(
1
1
)
58
oveq2
⊢
C
=
0
→
(
0
C
)
=
(
0
0
)
59
oveq1
⊢
C
=
0
→
C
+
1
=
0
+
1
60
59
4
eqtrdi
⊢
C
=
0
→
C
+
1
=
1
61
60
oveq2d
⊢
C
=
0
→
(
1
C
+
1
)
=
(
1
1
)
62
57
58
61
3eqtr4a
⊢
C
=
0
→
(
0
C
)
=
(
1
C
+
1
)
63
52
62
jaoi
⊢
C
∈
ℕ
∨
C
=
0
→
(
0
C
)
=
(
1
C
+
1
)
64
36
63
sylbi
⊢
C
∈
ℕ
0
→
(
0
C
)
=
(
1
C
+
1
)
65
35
64
eqtrd
⊢
C
∈
ℕ
0
→
∑
k
=
0
0
(
k
C
)
=
(
1
C
+
1
)
66
elnn0uz
⊢
n
∈
ℕ
0
↔
n
∈
ℤ
≥
0
67
66
biimpi
⊢
n
∈
ℕ
0
→
n
∈
ℤ
≥
0
68
67
adantr
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
n
∈
ℤ
≥
0
69
elfznn0
⊢
k
∈
0
…
n
+
1
→
k
∈
ℕ
0
70
69
adantl
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
k
∈
0
…
n
+
1
→
k
∈
ℕ
0
71
simplr
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
k
∈
0
…
n
+
1
→
C
∈
ℕ
0
72
71
nn0zd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
k
∈
0
…
n
+
1
→
C
∈
ℤ
73
bccl
⊢
k
∈
ℕ
0
∧
C
∈
ℤ
→
(
k
C
)
∈
ℕ
0
74
70
72
73
syl2anc
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
k
∈
0
…
n
+
1
→
(
k
C
)
∈
ℕ
0
75
74
nn0cnd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
k
∈
0
…
n
+
1
→
(
k
C
)
∈
ℂ
76
oveq1
⊢
k
=
n
+
1
→
(
k
C
)
=
(
n
+
1
C
)
77
68
75
76
fsump1
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
∑
k
=
0
n
+
1
(
k
C
)
=
∑
k
=
0
n
(
k
C
)
+
(
n
+
1
C
)
78
77
adantr
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
∑
k
=
0
n
+
1
(
k
C
)
=
∑
k
=
0
n
(
k
C
)
+
(
n
+
1
C
)
79
id
⊢
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
80
nn0cn
⊢
C
∈
ℕ
0
→
C
∈
ℂ
81
80
adantl
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
C
∈
ℂ
82
1cnd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
1
∈
ℂ
83
81
82
pncand
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
C
+
1
-
1
=
C
84
83
oveq2d
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
(
n
+
1
C
+
1
-
1
)
=
(
n
+
1
C
)
85
84
eqcomd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
(
n
+
1
C
)
=
(
n
+
1
C
+
1
-
1
)
86
79
85
oveqan12rd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
∑
k
=
0
n
(
k
C
)
+
(
n
+
1
C
)
=
(
n
+
1
C
+
1
)
+
(
n
+
1
C
+
1
-
1
)
87
peano2nn0
⊢
n
∈
ℕ
0
→
n
+
1
∈
ℕ
0
88
peano2nn0
⊢
C
∈
ℕ
0
→
C
+
1
∈
ℕ
0
89
88
nn0zd
⊢
C
∈
ℕ
0
→
C
+
1
∈
ℤ
90
bcpasc
⊢
n
+
1
∈
ℕ
0
∧
C
+
1
∈
ℤ
→
(
n
+
1
C
+
1
)
+
(
n
+
1
C
+
1
-
1
)
=
(
n
+
1
+
1
C
+
1
)
91
87
89
90
syl2an
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
→
(
n
+
1
C
+
1
)
+
(
n
+
1
C
+
1
-
1
)
=
(
n
+
1
+
1
C
+
1
)
92
91
adantr
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
(
n
+
1
C
+
1
)
+
(
n
+
1
C
+
1
-
1
)
=
(
n
+
1
+
1
C
+
1
)
93
78
86
92
3eqtrd
⊢
n
∈
ℕ
0
∧
C
∈
ℕ
0
∧
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
∑
k
=
0
n
+
1
(
k
C
)
=
(
n
+
1
+
1
C
+
1
)
94
93
exp31
⊢
n
∈
ℕ
0
→
C
∈
ℕ
0
→
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
∑
k
=
0
n
+
1
(
k
C
)
=
(
n
+
1
+
1
C
+
1
)
95
94
a2d
⊢
n
∈
ℕ
0
→
C
∈
ℕ
0
→
∑
k
=
0
n
(
k
C
)
=
(
n
+
1
C
+
1
)
→
C
∈
ℕ
0
→
∑
k
=
0
n
+
1
(
k
C
)
=
(
n
+
1
+
1
C
+
1
)
96
8
14
20
26
65
95
nn0ind
⊢
N
∈
ℕ
0
→
C
∈
ℕ
0
→
∑
k
=
0
N
(
k
C
)
=
(
N
+
1
C
+
1
)
97
96
imp
⊢
N
∈
ℕ
0
∧
C
∈
ℕ
0
→
∑
k
=
0
N
(
k
C
)
=
(
N
+
1
C
+
1
)