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 = fld 𝑠
qabsabv.a A = AbsVal Q
padic.j J = q x if x = 0 0 q q pCnt x
ostth.k K = x if x = 0 0 1
ostth.1 φ F A
ostth1.2 φ n ¬ 1 < F n
ostth1.3 φ n ¬ F n < 1
Assertion ostth1 φ F = K

Proof

Step Hyp Ref Expression
1 qrng.q Q = fld 𝑠
2 qabsabv.a A = AbsVal Q
3 padic.j J = q x if x = 0 0 q q pCnt x
4 ostth.k K = x if x = 0 0 1
5 ostth.1 φ F A
6 ostth1.2 φ n ¬ 1 < F n
7 ostth1.3 φ n ¬ F n < 1
8 1 qdrng Q DivRing
9 1 qrngbas = 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 ¬ F n < 1
14 prmnn n n
15 6 r19.21bi φ n ¬ 1 < F n
16 14 15 sylan2 φ n ¬ 1 < F n
17 nnq n n
18 14 17 syl n n
19 2 9 abvcl F A n F n
20 5 18 19 syl2an φ n F n
21 1re 1
22 lttri3 F n 1 F n = 1 ¬ F n < 1 ¬ 1 < F n
23 20 21 22 sylancl φ n F n = 1 ¬ F n < 1 ¬ 1 < F n
24 13 16 23 mpbir2and φ n F n = 1
25 14 adantl φ 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 K n = if n = 0 0 1
32 17 31 syl n K n = if n = 0 0 1
33 nnne0 n n 0
34 33 neneqd n ¬ n = 0
35 34 iffalsed n if n = 0 0 1 = 1
36 32 35 eqtrd n K n = 1
37 25 36 syl φ n K n = 1
38 24 37 eqtr4d φ n F n = K n
39 1 2 5 12 38 ostthlem2 φ F = K