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abelthlem7a
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abelthlem7
Metamath Proof Explorer
Ascii
Unicode
Theorem
abelthlem7a
Description:
Lemma for
abelth
.
(Contributed by
Mario Carneiro
, 8-May-2015)
Ref
Expression
Hypotheses
abelth.1
⊢
φ
→
A
:
ℕ
0
⟶
ℂ
abelth.2
⊢
φ
→
seq
0
+
A
∈
dom
⁡
⇝
abelth.3
⊢
φ
→
M
∈
ℝ
abelth.4
⊢
φ
→
0
≤
M
abelth.5
⊢
S
=
z
∈
ℂ
|
1
−
z
≤
M
⁢
1
−
z
abelth.6
⊢
F
=
x
∈
S
⟼
∑
n
∈
ℕ
0
A
⁡
n
⁢
x
n
abelth.7
⊢
φ
→
seq
0
+
A
⇝
0
abelthlem6.1
⊢
φ
→
X
∈
S
∖
1
Assertion
abelthlem7a
⊢
φ
→
X
∈
ℂ
∧
1
−
X
≤
M
⁢
1
−
X
Proof
Step
Hyp
Ref
Expression
1
abelth.1
⊢
φ
→
A
:
ℕ
0
⟶
ℂ
2
abelth.2
⊢
φ
→
seq
0
+
A
∈
dom
⁡
⇝
3
abelth.3
⊢
φ
→
M
∈
ℝ
4
abelth.4
⊢
φ
→
0
≤
M
5
abelth.5
⊢
S
=
z
∈
ℂ
|
1
−
z
≤
M
⁢
1
−
z
6
abelth.6
⊢
F
=
x
∈
S
⟼
∑
n
∈
ℕ
0
A
⁡
n
⁢
x
n
7
abelth.7
⊢
φ
→
seq
0
+
A
⇝
0
8
abelthlem6.1
⊢
φ
→
X
∈
S
∖
1
9
8
eldifad
⊢
φ
→
X
∈
S
10
oveq2
⊢
z
=
X
→
1
−
z
=
1
−
X
11
10
fveq2d
⊢
z
=
X
→
1
−
z
=
1
−
X
12
fveq2
⊢
z
=
X
→
z
=
X
13
12
oveq2d
⊢
z
=
X
→
1
−
z
=
1
−
X
14
13
oveq2d
⊢
z
=
X
→
M
⁢
1
−
z
=
M
⁢
1
−
X
15
11
14
breq12d
⊢
z
=
X
→
1
−
z
≤
M
⁢
1
−
z
↔
1
−
X
≤
M
⁢
1
−
X
16
15
5
elrab2
⊢
X
∈
S
↔
X
∈
ℂ
∧
1
−
X
≤
M
⁢
1
−
X
17
9
16
sylib
⊢
φ
→
X
∈
ℂ
∧
1
−
X
≤
M
⁢
1
−
X