Metamath Proof Explorer


Theorem ostth1

Description: - Lemma for ostth : trivial case. (Not that the proof is trivial, but that we are proving that the function is trivial.) If F is equal to 1 on the primes, then by complete induction and the multiplicative property abvmul of the absolute value, F is equal to 1 on all the integers, and ostthlem1 extends this to the other rational numbers. (Contributed by Mario Carneiro, 10-Sep-2014)

Ref Expression
Hypotheses qrng.q ⊢ Q = ℂ fld ↾ 𝑠 ℚ
qabsabv.a ⊢ A = AbsVal ⁡ Q
padic.j ⊢ J = q ∈ ℙ ⟼ x ∈ ℚ ⟼ if x = 0 0 q − q pCnt x
ostth.k ⊢ K = x ∈ ℚ ⟼ if x = 0 0 1
ostth.1 ⊢ φ → F ∈ A
ostth1.2 ⊢ φ → ∀ n ∈ ℕ ¬ 1 < F ⁡ n
ostth1.3 ⊢ φ → ∀ n ∈ ℙ ¬ F ⁡ n < 1
Assertion ostth1 ⊢ φ → F = K

Proof

Step Hyp Ref Expression
1 qrng.q ⊢ Q = ℂ fld ↾ 𝑠 ℚ
2 qabsabv.a ⊢ A = AbsVal ⁡ Q
3 padic.j ⊢ J = q ∈ ℙ ⟼ x ∈ ℚ ⟼ if x = 0 0 q − q pCnt x
4 ostth.k ⊢ K = x ∈ ℚ ⟼ if x = 0 0 1
5 ostth.1 ⊢ φ → F ∈ A
6 ostth1.2 ⊢ φ → ∀ n ∈ ℕ ¬ 1 < F ⁡ n
7 ostth1.3 ⊢ φ → ∀ n ∈ ℙ ¬ F ⁡ n < 1
8 1 qdrng ⊢ Q ∈ DivRing
9 1 qrngbas ⊢ ℚ = Base Q
10 1 qrng0 ⊢ 0 = 0 Q
11 2 9 10 4 abvtriv ⊢ Q ∈ DivRing → K ∈ A
12 8 11 mp1i ⊢ φ → K ∈ A
13 7 r19.21bi ⊢ φ ∧ n ∈ ℙ → ¬ F ⁡ n < 1
14 prmnn ⊢ n ∈ ℙ → n ∈ ℕ
15 6 r19.21bi ⊢ φ ∧ n ∈ ℕ → ¬ 1 < F ⁡ n
16 14 15 sylan2 ⊢ φ ∧ n ∈ ℙ → ¬ 1 < F ⁡ n
17 nnq ⊢ n ∈ ℕ → n ∈ ℚ
18 14 17 syl ⊢ n ∈ ℙ → n ∈ ℚ
19 2 9 abvcl ⊢ F ∈ A ∧ n ∈ ℚ → F ⁡ n ∈ ℝ
20 5 18 19 syl2an ⊢ φ ∧ n ∈ ℙ → F ⁡ n ∈ ℝ
21 1re ⊢ 1 ∈ ℝ
22 lttri3 ⊢ F ⁡ n ∈ ℝ ∧ 1 ∈ ℝ → F ⁡ n = 1 ↔ ¬ F ⁡ n < 1 ∧ ¬ 1 < F ⁡ n
23 20 21 22 sylancl ⊢ φ ∧ n ∈ ℙ → F ⁡ n = 1 ↔ ¬ F ⁡ n < 1 ∧ ¬ 1 < F ⁡ n
24 13 16 23 mpbir2and ⊢ φ ∧ n ∈ ℙ → F ⁡ n = 1
25 14 adantl ⊢ φ ∧ n ∈ ℙ → n ∈ ℕ
26 eqeq1 ⊢ x = n → x = 0 ↔ n = 0
27 26 ifbid ⊢ x = n → if x = 0 0 1 = if n = 0 0 1
28 c0ex ⊢ 0 ∈ V
29 1ex ⊢ 1 ∈ V
30 28 29 ifex ⊢ if n = 0 0 1 ∈ V
31 27 4 30 fvmpt ⊢ n ∈ ℚ → K ⁡ n = if n = 0 0 1
32 17 31 syl ⊢ n ∈ ℕ → K ⁡ n = if n = 0 0 1
33 nnne0 ⊢ n ∈ ℕ → n ≠ 0
34 33 neneqd ⊢ n ∈ ℕ → ¬ n = 0
35 34 iffalsed ⊢ n ∈ ℕ → if n = 0 0 1 = 1
36 32 35 eqtrd ⊢ n ∈ ℕ → K ⁡ n = 1
37 25 36 syl ⊢ φ ∧ n ∈ ℙ → K ⁡ n = 1
38 24 37 eqtr4d ⊢ φ ∧ n ∈ ℙ → F ⁡ n = K ⁡ n
39 1 2 5 12 38 ostthlem2 ⊢ φ → F = K