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BASIC REAL AND COMPLEX FUNCTIONS
Basic number theory
Gauss' Lemma
gausslemma2dlem4
Next ⟩
gausslemma2dlem5a
Metamath Proof Explorer
Ascii
Unicode
Theorem
gausslemma2dlem4
Description:
Lemma 4 for
gausslemma2d
.
(Contributed by
AV
, 16-Jun-2021)
Ref
Expression
Hypotheses
gausslemma2d.p
⊢
φ
→
P
∈
ℙ
∖
2
gausslemma2d.h
⊢
H
=
P
−
1
2
gausslemma2d.r
⊢
R
=
x
∈
1
…
H
⟼
if
x
⋅
2
<
P
2
x
⋅
2
P
−
x
⋅
2
gausslemma2d.m
⊢
M
=
P
4
Assertion
gausslemma2dlem4
⊢
φ
→
H
!
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
Proof
Step
Hyp
Ref
Expression
1
gausslemma2d.p
⊢
φ
→
P
∈
ℙ
∖
2
2
gausslemma2d.h
⊢
H
=
P
−
1
2
3
gausslemma2d.r
⊢
R
=
x
∈
1
…
H
⟼
if
x
⋅
2
<
P
2
x
⋅
2
P
−
x
⋅
2
4
gausslemma2d.m
⊢
M
=
P
4
5
1
2
3
gausslemma2dlem1
⊢
φ
→
H
!
=
∏
k
=
1
H
R
k
6
eldif
⊢
P
∈
ℙ
∖
2
↔
P
∈
ℙ
∧
¬
P
∈
2
7
prm23ge5
⊢
P
∈
ℙ
→
P
=
2
∨
P
=
3
∨
P
∈
ℤ
≥
5
8
eleq1
⊢
P
=
2
→
P
∈
2
↔
2
∈
2
9
8
notbid
⊢
P
=
2
→
¬
P
∈
2
↔
¬
2
∈
2
10
2ex
⊢
2
∈
V
11
10
snid
⊢
2
∈
2
12
11
2a1i
⊢
P
=
2
→
∏
k
=
1
H
R
k
≠
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
→
2
∈
2
13
12
necon1bd
⊢
P
=
2
→
¬
2
∈
2
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
14
13
a1dd
⊢
P
=
2
→
¬
2
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
15
9
14
sylbid
⊢
P
=
2
→
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
16
3lt4
⊢
3
<
4
17
breq1
⊢
P
=
3
→
P
<
4
↔
3
<
4
18
16
17
mpbiri
⊢
P
=
3
→
P
<
4
19
3nn0
⊢
3
∈
ℕ
0
20
eleq1
⊢
P
=
3
→
P
∈
ℕ
0
↔
3
∈
ℕ
0
21
19
20
mpbiri
⊢
P
=
3
→
P
∈
ℕ
0
22
4nn
⊢
4
∈
ℕ
23
divfl0
⊢
P
∈
ℕ
0
∧
4
∈
ℕ
→
P
<
4
↔
P
4
=
0
24
21
22
23
sylancl
⊢
P
=
3
→
P
<
4
↔
P
4
=
0
25
18
24
mpbid
⊢
P
=
3
→
P
4
=
0
26
4
25
eqtrid
⊢
P
=
3
→
M
=
0
27
oveq2
⊢
M
=
0
→
1
…
M
=
1
…
0
28
27
adantr
⊢
M
=
0
∧
φ
→
1
…
M
=
1
…
0
29
fz10
⊢
1
…
0
=
∅
30
28
29
eqtrdi
⊢
M
=
0
∧
φ
→
1
…
M
=
∅
31
30
prodeq1d
⊢
M
=
0
∧
φ
→
∏
k
=
1
M
R
k
=
∏
k
∈
∅
R
k
32
prod0
⊢
∏
k
∈
∅
R
k
=
1
33
31
32
eqtrdi
⊢
M
=
0
∧
φ
→
∏
k
=
1
M
R
k
=
1
34
oveq1
⊢
M
=
0
→
M
+
1
=
0
+
1
35
34
adantr
⊢
M
=
0
∧
φ
→
M
+
1
=
0
+
1
36
0p1e1
⊢
0
+
1
=
1
37
35
36
eqtrdi
⊢
M
=
0
∧
φ
→
M
+
1
=
1
38
37
oveq1d
⊢
M
=
0
∧
φ
→
M
+
1
…
H
=
1
…
H
39
38
prodeq1d
⊢
M
=
0
∧
φ
→
∏
k
=
M
+
1
H
R
k
=
∏
k
=
1
H
R
k
40
33
39
oveq12d
⊢
M
=
0
∧
φ
→
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
=
1
∏
k
=
1
H
R
k
41
fzfid
⊢
M
=
0
∧
φ
→
1
…
H
∈
Fin
42
oveq1
⊢
x
=
k
→
x
⋅
2
=
k
⋅
2
43
42
breq1d
⊢
x
=
k
→
x
⋅
2
<
P
2
↔
k
⋅
2
<
P
2
44
42
oveq2d
⊢
x
=
k
→
P
−
x
⋅
2
=
P
−
k
⋅
2
45
43
42
44
ifbieq12d
⊢
x
=
k
→
if
x
⋅
2
<
P
2
x
⋅
2
P
−
x
⋅
2
=
if
k
⋅
2
<
P
2
k
⋅
2
P
−
k
⋅
2
46
simpr
⊢
φ
∧
k
∈
1
…
H
→
k
∈
1
…
H
47
elfzelz
⊢
k
∈
1
…
H
→
k
∈
ℤ
48
47
zcnd
⊢
k
∈
1
…
H
→
k
∈
ℂ
49
2cnd
⊢
k
∈
1
…
H
→
2
∈
ℂ
50
48
49
mulcld
⊢
k
∈
1
…
H
→
k
⋅
2
∈
ℂ
51
50
adantl
⊢
φ
∧
k
∈
1
…
H
→
k
⋅
2
∈
ℂ
52
eldifi
⊢
P
∈
ℙ
∖
2
→
P
∈
ℙ
53
prmz
⊢
P
∈
ℙ
→
P
∈
ℤ
54
53
zcnd
⊢
P
∈
ℙ
→
P
∈
ℂ
55
1
52
54
3syl
⊢
φ
→
P
∈
ℂ
56
55
adantr
⊢
φ
∧
k
∈
1
…
H
→
P
∈
ℂ
57
56
51
subcld
⊢
φ
∧
k
∈
1
…
H
→
P
−
k
⋅
2
∈
ℂ
58
51
57
ifcld
⊢
φ
∧
k
∈
1
…
H
→
if
k
⋅
2
<
P
2
k
⋅
2
P
−
k
⋅
2
∈
ℂ
59
3
45
46
58
fvmptd3
⊢
φ
∧
k
∈
1
…
H
→
R
k
=
if
k
⋅
2
<
P
2
k
⋅
2
P
−
k
⋅
2
60
59
58
eqeltrd
⊢
φ
∧
k
∈
1
…
H
→
R
k
∈
ℂ
61
60
adantll
⊢
M
=
0
∧
φ
∧
k
∈
1
…
H
→
R
k
∈
ℂ
62
41
61
fprodcl
⊢
M
=
0
∧
φ
→
∏
k
=
1
H
R
k
∈
ℂ
63
62
mullidd
⊢
M
=
0
∧
φ
→
1
∏
k
=
1
H
R
k
=
∏
k
=
1
H
R
k
64
40
63
eqtr2d
⊢
M
=
0
∧
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
65
64
ex
⊢
M
=
0
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
66
26
65
syl
⊢
P
=
3
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
67
66
a1d
⊢
P
=
3
→
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
68
1
4
gausslemma2dlem0d
⊢
φ
→
M
∈
ℕ
0
69
68
nn0red
⊢
φ
→
M
∈
ℝ
70
69
ltp1d
⊢
φ
→
M
<
M
+
1
71
fzdisj
⊢
M
<
M
+
1
→
1
…
M
∩
M
+
1
…
H
=
∅
72
70
71
syl
⊢
φ
→
1
…
M
∩
M
+
1
…
H
=
∅
73
72
adantl
⊢
P
∈
ℤ
≥
5
∧
φ
→
1
…
M
∩
M
+
1
…
H
=
∅
74
eluzelre
⊢
P
∈
ℤ
≥
5
→
P
∈
ℝ
75
4re
⊢
4
∈
ℝ
76
75
a1i
⊢
P
∈
ℤ
≥
5
→
4
∈
ℝ
77
4ne0
⊢
4
≠
0
78
77
a1i
⊢
P
∈
ℤ
≥
5
→
4
≠
0
79
74
76
78
redivcld
⊢
P
∈
ℤ
≥
5
→
P
4
∈
ℝ
80
79
flcld
⊢
P
∈
ℤ
≥
5
→
P
4
∈
ℤ
81
nnrp
⊢
4
∈
ℕ
→
4
∈
ℝ
+
82
22
81
ax-mp
⊢
4
∈
ℝ
+
83
eluz2
⊢
P
∈
ℤ
≥
5
↔
5
∈
ℤ
∧
P
∈
ℤ
∧
5
≤
P
84
4lt5
⊢
4
<
5
85
5re
⊢
5
∈
ℝ
86
85
a1i
⊢
5
∈
ℤ
∧
P
∈
ℤ
→
5
∈
ℝ
87
zre
⊢
P
∈
ℤ
→
P
∈
ℝ
88
87
adantl
⊢
5
∈
ℤ
∧
P
∈
ℤ
→
P
∈
ℝ
89
ltleletr
⊢
4
∈
ℝ
∧
5
∈
ℝ
∧
P
∈
ℝ
→
4
<
5
∧
5
≤
P
→
4
≤
P
90
75
86
88
89
mp3an2i
⊢
5
∈
ℤ
∧
P
∈
ℤ
→
4
<
5
∧
5
≤
P
→
4
≤
P
91
84
90
mpani
⊢
5
∈
ℤ
∧
P
∈
ℤ
→
5
≤
P
→
4
≤
P
92
91
3impia
⊢
5
∈
ℤ
∧
P
∈
ℤ
∧
5
≤
P
→
4
≤
P
93
83
92
sylbi
⊢
P
∈
ℤ
≥
5
→
4
≤
P
94
divge1
⊢
4
∈
ℝ
+
∧
P
∈
ℝ
∧
4
≤
P
→
1
≤
P
4
95
82
74
93
94
mp3an2i
⊢
P
∈
ℤ
≥
5
→
1
≤
P
4
96
1zzd
⊢
P
∈
ℤ
≥
5
→
1
∈
ℤ
97
flge
⊢
P
4
∈
ℝ
∧
1
∈
ℤ
→
1
≤
P
4
↔
1
≤
P
4
98
79
96
97
syl2anc
⊢
P
∈
ℤ
≥
5
→
1
≤
P
4
↔
1
≤
P
4
99
95
98
mpbid
⊢
P
∈
ℤ
≥
5
→
1
≤
P
4
100
elnnz1
⊢
P
4
∈
ℕ
↔
P
4
∈
ℤ
∧
1
≤
P
4
101
80
99
100
sylanbrc
⊢
P
∈
ℤ
≥
5
→
P
4
∈
ℕ
102
101
adantl
⊢
P
∈
ℙ
∖
2
∧
P
∈
ℤ
≥
5
→
P
4
∈
ℕ
103
oddprm
⊢
P
∈
ℙ
∖
2
→
P
−
1
2
∈
ℕ
104
103
adantr
⊢
P
∈
ℙ
∖
2
∧
P
∈
ℤ
≥
5
→
P
−
1
2
∈
ℕ
105
prmuz2
⊢
P
∈
ℙ
→
P
∈
ℤ
≥
2
106
52
105
syl
⊢
P
∈
ℙ
∖
2
→
P
∈
ℤ
≥
2
107
106
adantr
⊢
P
∈
ℙ
∖
2
∧
P
∈
ℤ
≥
5
→
P
∈
ℤ
≥
2
108
fldiv4lem1div2uz2
⊢
P
∈
ℤ
≥
2
→
P
4
≤
P
−
1
2
109
107
108
syl
⊢
P
∈
ℙ
∖
2
∧
P
∈
ℤ
≥
5
→
P
4
≤
P
−
1
2
110
102
104
109
3jca
⊢
P
∈
ℙ
∖
2
∧
P
∈
ℤ
≥
5
→
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
111
110
ex
⊢
P
∈
ℙ
∖
2
→
P
∈
ℤ
≥
5
→
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
112
1
111
syl
⊢
φ
→
P
∈
ℤ
≥
5
→
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
113
112
impcom
⊢
P
∈
ℤ
≥
5
∧
φ
→
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
114
2
oveq2i
⊢
1
…
H
=
1
…
P
−
1
2
115
4
114
eleq12i
⊢
M
∈
1
…
H
↔
P
4
∈
1
…
P
−
1
2
116
elfz1b
⊢
P
4
∈
1
…
P
−
1
2
↔
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
117
115
116
bitri
⊢
M
∈
1
…
H
↔
P
4
∈
ℕ
∧
P
−
1
2
∈
ℕ
∧
P
4
≤
P
−
1
2
118
113
117
sylibr
⊢
P
∈
ℤ
≥
5
∧
φ
→
M
∈
1
…
H
119
fzsplit
⊢
M
∈
1
…
H
→
1
…
H
=
1
…
M
∪
M
+
1
…
H
120
118
119
syl
⊢
P
∈
ℤ
≥
5
∧
φ
→
1
…
H
=
1
…
M
∪
M
+
1
…
H
121
fzfid
⊢
P
∈
ℤ
≥
5
∧
φ
→
1
…
H
∈
Fin
122
60
adantll
⊢
P
∈
ℤ
≥
5
∧
φ
∧
k
∈
1
…
H
→
R
k
∈
ℂ
123
73
120
121
122
fprodsplit
⊢
P
∈
ℤ
≥
5
∧
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
124
123
ex
⊢
P
∈
ℤ
≥
5
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
125
124
a1d
⊢
P
∈
ℤ
≥
5
→
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
126
15
67
125
3jaoi
⊢
P
=
2
∨
P
=
3
∨
P
∈
ℤ
≥
5
→
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
127
7
126
syl
⊢
P
∈
ℙ
→
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
128
127
imp
⊢
P
∈
ℙ
∧
¬
P
∈
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
129
6
128
sylbi
⊢
P
∈
ℙ
∖
2
→
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
130
1
129
mpcom
⊢
φ
→
∏
k
=
1
H
R
k
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k
131
5
130
eqtrd
⊢
φ
→
H
!
=
∏
k
=
1
M
R
k
∏
k
=
M
+
1
H
R
k