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=fld𝑠
qabsabv.a A=AbsValQ
padic.j J=qxifx=00qqpCntx
ostth.k K=xifx=001
ostth.1 φFA
ostth1.2 φn¬1<Fn
ostth1.3 φn¬Fn<1
Assertion ostth1 φF=K

Proof

Step Hyp Ref Expression
1 qrng.q Q=fld𝑠
2 qabsabv.a A=AbsValQ
3 padic.j J=qxifx=00qqpCntx
4 ostth.k K=xifx=001
5 ostth.1 φFA
6 ostth1.2 φn¬1<Fn
7 ostth1.3 φn¬Fn<1
8 1 qdrng QDivRing
9 1 qrngbas =BaseQ
10 1 qrng0 0=0Q
11 2 9 10 4 abvtriv QDivRingKA
12 8 11 mp1i φKA
13 7 r19.21bi φn¬Fn<1
14 prmnn nn
15 6 r19.21bi φn¬1<Fn
16 14 15 sylan2 φn¬1<Fn
17 nnq nn
18 14 17 syl nn
19 2 9 abvcl FAnFn
20 5 18 19 syl2an φnFn
21 1re 1
22 lttri3 Fn1Fn=1¬Fn<1¬1<Fn
23 20 21 22 sylancl φnFn=1¬Fn<1¬1<Fn
24 13 16 23 mpbir2and φnFn=1
25 14 adantl φnn
26 eqeq1 x=nx=0n=0
27 26 ifbid x=nifx=001=ifn=001
28 c0ex 0V
29 1ex 1V
30 28 29 ifex ifn=001V
31 27 4 30 fvmpt nKn=ifn=001
32 17 31 syl nKn=ifn=001
33 nnne0 nn0
34 33 neneqd n¬n=0
35 34 iffalsed nifn=001=1
36 32 35 eqtrd nKn=1
37 25 36 syl φnKn=1
38 24 37 eqtr4d φnFn=Kn
39 1 2 5 12 38 ostthlem2 φF=K