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Mathbox for Stanislas Polu
inductionexd
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Theorem
inductionexd
Description:
Simple induction example.
(Contributed by
Stanislas Polu
, 9-Mar-2020)
Ref
Expression
Assertion
inductionexd
⊢
N
∈
ℕ
→
3
∥
4
N
+
5
Proof
Step
Hyp
Ref
Expression
1
oveq2
⊢
k
=
1
→
4
k
=
4
1
2
1
oveq1d
⊢
k
=
1
→
4
k
+
5
=
4
1
+
5
3
2
breq2d
⊢
k
=
1
→
3
∥
4
k
+
5
↔
3
∥
4
1
+
5
4
oveq2
⊢
k
=
n
→
4
k
=
4
n
5
4
oveq1d
⊢
k
=
n
→
4
k
+
5
=
4
n
+
5
6
5
breq2d
⊢
k
=
n
→
3
∥
4
k
+
5
↔
3
∥
4
n
+
5
7
oveq2
⊢
k
=
n
+
1
→
4
k
=
4
n
+
1
8
7
oveq1d
⊢
k
=
n
+
1
→
4
k
+
5
=
4
n
+
1
+
5
9
8
breq2d
⊢
k
=
n
+
1
→
3
∥
4
k
+
5
↔
3
∥
4
n
+
1
+
5
10
oveq2
⊢
k
=
N
→
4
k
=
4
N
11
10
oveq1d
⊢
k
=
N
→
4
k
+
5
=
4
N
+
5
12
11
breq2d
⊢
k
=
N
→
3
∥
4
k
+
5
↔
3
∥
4
N
+
5
13
3z
⊢
3
∈
ℤ
14
4z
⊢
4
∈
ℤ
15
1nn0
⊢
1
∈
ℕ
0
16
zexpcl
⊢
4
∈
ℤ
∧
1
∈
ℕ
0
→
4
1
∈
ℤ
17
14
15
16
mp2an
⊢
4
1
∈
ℤ
18
5nn
⊢
5
∈
ℕ
19
18
nnzi
⊢
5
∈
ℤ
20
zaddcl
⊢
4
1
∈
ℤ
∧
5
∈
ℤ
→
4
1
+
5
∈
ℤ
21
17
19
20
mp2an
⊢
4
1
+
5
∈
ℤ
22
13
13
21
3pm3.2i
⊢
3
∈
ℤ
∧
3
∈
ℤ
∧
4
1
+
5
∈
ℤ
23
3t3e9
⊢
3
⋅
3
=
9
24
4nn0
⊢
4
∈
ℕ
0
25
24
numexp1
⊢
4
1
=
4
26
25
oveq1i
⊢
4
1
+
5
=
4
+
5
27
5cn
⊢
5
∈
ℂ
28
4cn
⊢
4
∈
ℂ
29
5p4e9
⊢
5
+
4
=
9
30
27
28
29
addcomli
⊢
4
+
5
=
9
31
26
30
eqtri
⊢
4
1
+
5
=
9
32
23
31
eqtr4i
⊢
3
⋅
3
=
4
1
+
5
33
dvds0lem
⊢
3
∈
ℤ
∧
3
∈
ℤ
∧
4
1
+
5
∈
ℤ
∧
3
⋅
3
=
4
1
+
5
→
3
∥
4
1
+
5
34
22
32
33
mp2an
⊢
3
∥
4
1
+
5
35
13
a1i
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∈
ℤ
36
4nn
⊢
4
∈
ℕ
37
36
a1i
⊢
n
∈
ℕ
→
4
∈
ℕ
38
nnnn0
⊢
n
∈
ℕ
→
n
∈
ℕ
0
39
37
38
nnexpcld
⊢
n
∈
ℕ
→
4
n
∈
ℕ
40
39
nnzd
⊢
n
∈
ℕ
→
4
n
∈
ℤ
41
40
adantr
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
4
n
∈
ℤ
42
19
a1i
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
5
∈
ℤ
43
41
42
zaddcld
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
4
n
+
5
∈
ℤ
44
14
a1i
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
4
∈
ℤ
45
43
44
zmulcld
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
4
n
+
5
⋅
4
∈
ℤ
46
35
42
zmulcld
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
⋅
5
∈
ℤ
47
simpr
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∥
4
n
+
5
48
35
43
44
47
dvdsmultr1d
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∥
4
n
+
5
⋅
4
49
dvdsmul1
⊢
3
∈
ℤ
∧
5
∈
ℤ
→
3
∥
3
⋅
5
50
13
19
49
mp2an
⊢
3
∥
3
⋅
5
51
50
a1i
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∥
3
⋅
5
52
35
45
46
48
51
dvds2subd
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∥
4
n
+
5
⋅
4
−
3
⋅
5
53
39
nncnd
⊢
n
∈
ℕ
→
4
n
∈
ℂ
54
27
a1i
⊢
n
∈
ℕ
→
5
∈
ℂ
55
28
a1i
⊢
n
∈
ℕ
→
4
∈
ℂ
56
53
54
55
adddird
⊢
n
∈
ℕ
→
4
n
+
5
⋅
4
=
4
n
⋅
4
+
5
⋅
4
57
56
oveq1d
⊢
n
∈
ℕ
→
4
n
+
5
⋅
4
−
15
=
4
n
⋅
4
+
5
⋅
4
-
15
58
3cn
⊢
3
∈
ℂ
59
5t3e15
⊢
5
⋅
3
=
15
60
27
58
59
mulcomli
⊢
3
⋅
5
=
15
61
60
a1i
⊢
n
∈
ℕ
→
3
⋅
5
=
15
62
61
oveq2d
⊢
n
∈
ℕ
→
4
n
+
5
⋅
4
−
3
⋅
5
=
4
n
+
5
⋅
4
−
15
63
55
38
expp1d
⊢
n
∈
ℕ
→
4
n
+
1
=
4
n
⋅
4
64
ax-1cn
⊢
1
∈
ℂ
65
3p1e4
⊢
3
+
1
=
4
66
58
64
65
addcomli
⊢
1
+
3
=
4
67
66
eqcomi
⊢
4
=
1
+
3
68
67
oveq1i
⊢
4
−
3
=
1
+
3
-
3
69
64
58
pncan3oi
⊢
1
+
3
-
3
=
1
70
68
69
eqtri
⊢
4
−
3
=
1
71
70
oveq2i
⊢
5
4
−
3
=
5
⋅
1
72
27
28
58
subdii
⊢
5
4
−
3
=
5
⋅
4
−
5
⋅
3
73
27
mulridi
⊢
5
⋅
1
=
5
74
71
72
73
3eqtr3ri
⊢
5
=
5
⋅
4
−
5
⋅
3
75
59
eqcomi
⊢
15
=
5
⋅
3
76
75
oveq2i
⊢
5
⋅
4
−
15
=
5
⋅
4
−
5
⋅
3
77
74
76
eqtr4i
⊢
5
=
5
⋅
4
−
15
78
77
a1i
⊢
n
∈
ℕ
→
5
=
5
⋅
4
−
15
79
63
78
oveq12d
⊢
n
∈
ℕ
→
4
n
+
1
+
5
=
4
n
⋅
4
+
5
⋅
4
-
15
80
53
55
mulcld
⊢
n
∈
ℕ
→
4
n
⋅
4
∈
ℂ
81
54
55
mulcld
⊢
n
∈
ℕ
→
5
⋅
4
∈
ℂ
82
5nn0
⊢
5
∈
ℕ
0
83
15
82
deccl
⊢
15
∈
ℕ
0
84
83
nn0cni
⊢
15
∈
ℂ
85
84
a1i
⊢
n
∈
ℕ
→
15
∈
ℂ
86
80
81
85
addsubassd
⊢
n
∈
ℕ
→
4
n
⋅
4
+
5
⋅
4
-
15
=
4
n
⋅
4
+
5
⋅
4
-
15
87
79
86
eqtr4d
⊢
n
∈
ℕ
→
4
n
+
1
+
5
=
4
n
⋅
4
+
5
⋅
4
-
15
88
57
62
87
3eqtr4rd
⊢
n
∈
ℕ
→
4
n
+
1
+
5
=
4
n
+
5
⋅
4
−
3
⋅
5
89
88
adantr
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
4
n
+
1
+
5
=
4
n
+
5
⋅
4
−
3
⋅
5
90
52
89
breqtrrd
⊢
n
∈
ℕ
∧
3
∥
4
n
+
5
→
3
∥
4
n
+
1
+
5
91
90
ex
⊢
n
∈
ℕ
→
3
∥
4
n
+
5
→
3
∥
4
n
+
1
+
5
92
3
6
9
12
34
91
nnind
⊢
N
∈
ℕ
→
3
∥
4
N
+
5