Database
BASIC REAL AND COMPLEX FUNCTIONS
Polynomials
The division algorithm for polynomials
quotcan
Next ⟩
vieta1lem1
Metamath Proof Explorer
Ascii
Unicode
Theorem
quotcan
Description:
Exact division with a multiple.
(Contributed by
Mario Carneiro
, 26-Jul-2014)
Ref
Expression
Hypothesis
quotcan.1
⊢
H
=
F
×
f
G
Assertion
quotcan
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
quot
G
=
F
Proof
Step
Hyp
Ref
Expression
1
quotcan.1
⊢
H
=
F
×
f
G
2
plyssc
⊢
Poly
S
⊆
Poly
ℂ
3
simp2
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
∈
Poly
S
4
2
3
sselid
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
∈
Poly
ℂ
5
simp1
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
∈
Poly
S
6
2
5
sselid
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
∈
Poly
ℂ
7
plymulcl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
→
F
×
f
G
∈
Poly
ℂ
8
1
7
eqeltrid
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
→
H
∈
Poly
ℂ
9
8
3adant3
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
∈
Poly
ℂ
10
simp3
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
≠
0
𝑝
11
quotcl2
⊢
H
∈
Poly
ℂ
∧
G
∈
Poly
ℂ
∧
G
≠
0
𝑝
→
H
quot
G
∈
Poly
ℂ
12
9
4
10
11
syl3anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
quot
G
∈
Poly
ℂ
13
plysubcl
⊢
F
∈
Poly
ℂ
∧
H
quot
G
∈
Poly
ℂ
→
F
−
f
H
quot
G
∈
Poly
ℂ
14
6
12
13
syl2anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
∈
Poly
ℂ
15
plymul0or
⊢
G
∈
Poly
ℂ
∧
F
−
f
H
quot
G
∈
Poly
ℂ
→
G
×
f
F
−
f
H
quot
G
=
0
𝑝
↔
G
=
0
𝑝
∨
F
−
f
H
quot
G
=
0
𝑝
16
4
14
15
syl2anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
×
f
F
−
f
H
quot
G
=
0
𝑝
↔
G
=
0
𝑝
∨
F
−
f
H
quot
G
=
0
𝑝
17
cnex
⊢
ℂ
∈
V
18
17
a1i
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
ℂ
∈
V
19
plyf
⊢
F
∈
Poly
S
→
F
:
ℂ
⟶
ℂ
20
5
19
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
:
ℂ
⟶
ℂ
21
plyf
⊢
G
∈
Poly
S
→
G
:
ℂ
⟶
ℂ
22
3
21
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
:
ℂ
⟶
ℂ
23
mulcom
⊢
x
∈
ℂ
∧
y
∈
ℂ
→
x
y
=
y
x
24
23
adantl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
∧
x
∈
ℂ
∧
y
∈
ℂ
→
x
y
=
y
x
25
18
20
22
24
caofcom
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
×
f
G
=
G
×
f
F
26
1
25
eqtrid
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
=
G
×
f
F
27
26
oveq1d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
G
×
f
F
−
f
G
×
f
H
quot
G
28
plyf
⊢
H
quot
G
∈
Poly
ℂ
→
H
quot
G
:
ℂ
⟶
ℂ
29
12
28
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
quot
G
:
ℂ
⟶
ℂ
30
subdi
⊢
x
∈
ℂ
∧
y
∈
ℂ
∧
z
∈
ℂ
→
x
y
−
z
=
x
y
−
x
z
31
30
adantl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
∧
x
∈
ℂ
∧
y
∈
ℂ
∧
z
∈
ℂ
→
x
y
−
z
=
x
y
−
x
z
32
18
22
20
29
31
caofdi
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
×
f
F
−
f
H
quot
G
=
G
×
f
F
−
f
G
×
f
H
quot
G
33
27
32
eqtr4d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
G
×
f
F
−
f
H
quot
G
34
33
eqeq1d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
0
𝑝
↔
G
×
f
F
−
f
H
quot
G
=
0
𝑝
35
10
neneqd
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
¬
G
=
0
𝑝
36
biorf
⊢
¬
G
=
0
𝑝
→
F
−
f
H
quot
G
=
0
𝑝
↔
G
=
0
𝑝
∨
F
−
f
H
quot
G
=
0
𝑝
37
35
36
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
=
0
𝑝
↔
G
=
0
𝑝
∨
F
−
f
H
quot
G
=
0
𝑝
38
16
34
37
3bitr4d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
0
𝑝
↔
F
−
f
H
quot
G
=
0
𝑝
39
38
biimpd
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
0
𝑝
→
F
−
f
H
quot
G
=
0
𝑝
40
eqid
⊢
deg
G
=
deg
G
41
eqid
⊢
deg
F
−
f
H
quot
G
=
deg
F
−
f
H
quot
G
42
40
41
dgrmul
⊢
G
∈
Poly
ℂ
∧
G
≠
0
𝑝
∧
F
−
f
H
quot
G
∈
Poly
ℂ
∧
F
−
f
H
quot
G
≠
0
𝑝
→
deg
G
×
f
F
−
f
H
quot
G
=
deg
G
+
deg
F
−
f
H
quot
G
43
42
expr
⊢
G
∈
Poly
ℂ
∧
G
≠
0
𝑝
∧
F
−
f
H
quot
G
∈
Poly
ℂ
→
F
−
f
H
quot
G
≠
0
𝑝
→
deg
G
×
f
F
−
f
H
quot
G
=
deg
G
+
deg
F
−
f
H
quot
G
44
4
10
14
43
syl21anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
≠
0
𝑝
→
deg
G
×
f
F
−
f
H
quot
G
=
deg
G
+
deg
F
−
f
H
quot
G
45
dgrcl
⊢
G
∈
Poly
S
→
deg
G
∈
ℕ
0
46
3
45
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
∈
ℕ
0
47
46
nn0red
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
∈
ℝ
48
dgrcl
⊢
F
−
f
H
quot
G
∈
Poly
ℂ
→
deg
F
−
f
H
quot
G
∈
ℕ
0
49
14
48
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
F
−
f
H
quot
G
∈
ℕ
0
50
nn0addge1
⊢
deg
G
∈
ℝ
∧
deg
F
−
f
H
quot
G
∈
ℕ
0
→
deg
G
≤
deg
G
+
deg
F
−
f
H
quot
G
51
47
49
50
syl2anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
≤
deg
G
+
deg
F
−
f
H
quot
G
52
breq2
⊢
deg
G
×
f
F
−
f
H
quot
G
=
deg
G
+
deg
F
−
f
H
quot
G
→
deg
G
≤
deg
G
×
f
F
−
f
H
quot
G
↔
deg
G
≤
deg
G
+
deg
F
−
f
H
quot
G
53
51
52
syl5ibrcom
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
×
f
F
−
f
H
quot
G
=
deg
G
+
deg
F
−
f
H
quot
G
→
deg
G
≤
deg
G
×
f
F
−
f
H
quot
G
54
44
53
syld
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
≠
0
𝑝
→
deg
G
≤
deg
G
×
f
F
−
f
H
quot
G
55
33
fveq2d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
H
−
f
G
×
f
H
quot
G
=
deg
G
×
f
F
−
f
H
quot
G
56
55
breq2d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
≤
deg
H
−
f
G
×
f
H
quot
G
↔
deg
G
≤
deg
G
×
f
F
−
f
H
quot
G
57
plymulcl
⊢
G
∈
Poly
ℂ
∧
H
quot
G
∈
Poly
ℂ
→
G
×
f
H
quot
G
∈
Poly
ℂ
58
4
12
57
syl2anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
G
×
f
H
quot
G
∈
Poly
ℂ
59
plysubcl
⊢
H
∈
Poly
ℂ
∧
G
×
f
H
quot
G
∈
Poly
ℂ
→
H
−
f
G
×
f
H
quot
G
∈
Poly
ℂ
60
9
58
59
syl2anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
∈
Poly
ℂ
61
dgrcl
⊢
H
−
f
G
×
f
H
quot
G
∈
Poly
ℂ
→
deg
H
−
f
G
×
f
H
quot
G
∈
ℕ
0
62
60
61
syl
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
H
−
f
G
×
f
H
quot
G
∈
ℕ
0
63
62
nn0red
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
H
−
f
G
×
f
H
quot
G
∈
ℝ
64
47
63
lenltd
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
≤
deg
H
−
f
G
×
f
H
quot
G
↔
¬
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
65
56
64
bitr3d
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
G
≤
deg
G
×
f
F
−
f
H
quot
G
↔
¬
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
66
54
65
sylibd
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
≠
0
𝑝
→
¬
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
67
66
necon4ad
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
→
F
−
f
H
quot
G
=
0
𝑝
68
eqid
⊢
H
−
f
G
×
f
H
quot
G
=
H
−
f
G
×
f
H
quot
G
69
68
quotdgr
⊢
H
∈
Poly
ℂ
∧
G
∈
Poly
ℂ
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
0
𝑝
∨
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
70
9
4
10
69
syl3anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
−
f
G
×
f
H
quot
G
=
0
𝑝
∨
deg
H
−
f
G
×
f
H
quot
G
<
deg
G
71
39
67
70
mpjaod
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
=
0
𝑝
72
df-0p
⊢
0
𝑝
=
ℂ
×
0
73
71
72
eqtrdi
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
=
ℂ
×
0
74
ofsubeq0
⊢
ℂ
∈
V
∧
F
:
ℂ
⟶
ℂ
∧
H
quot
G
:
ℂ
⟶
ℂ
→
F
−
f
H
quot
G
=
ℂ
×
0
↔
F
=
H
quot
G
75
18
20
29
74
syl3anc
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
−
f
H
quot
G
=
ℂ
×
0
↔
F
=
H
quot
G
76
73
75
mpbid
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
F
=
H
quot
G
77
76
eqcomd
⊢
F
∈
Poly
S
∧
G
∈
Poly
S
∧
G
≠
0
𝑝
→
H
quot
G
=
F