My discovery about mathematics

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Naoto Meguro. Amateur.
MSC 2010. Primary 03C55; Secondary 11E26.
Key Words and Phrases.

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Let X be countable and consistent axiom system of the mathematics like the following.
Let N",R",C" be the individual symbols which become N,R,C for the standard model of X.
Let d be a free individual symbol. X doesn't include d. Let s₁,s₂,…(∈C) be the zero points of ζ(s).
Set up that (s₁∈C") ∈X, (s₂∈C") ∈X,….
Theorem 1. X has a countable model M for which M |= (d∈N")∧(d≥n) for ∀n∈N.
Proof. X∪{d∈N", d≥1,d≥2,…,d≥m} has a model M_m for which
M_m |= (d=m+1)∧(N"=N)∧(R"=R)∧(C"=C) and is consistent for ∀m∈N.
So X∪{d∈N", d≥1, d≥2,…(infinitely)} is consistent and has a countable model M for which
M |= (d∈N")∧(d≥n) for ∀n∈N.♦
Let M be that in theorem 1. Let O be the object domain of M. Define M" and μ"(s) by
M" |= μ"(n)=μ(n) (Möbius function) when n∈N and M" |= μ"(n)=0 for the other cases.
and M |= p(c₁,…,c_k) ⇒ M" |= p(c₁,…,c_k). p(x₁,…,x_k) is an arbitrary predicate symbol. c_i∈O.
The object domain of M" is O. M" is a model of X. Put H={s∈C"| ¬(Re(s)≤0)}.
If s∈H and n∈N,M" |= |n^s|≥1. M" |= |∑_1≤m≤n μ(m)/m^s|≤1+1+…+1=n≤d.
In the standard mathematics,1/ζ(s)=∑_1≤n≤∞ μ(n)/n^s (¬(Re(s)≤1)).
So 1/ζ(s) has no pole on H for M". ζ(s) has no zero point on H for M". If ∃s_m∈H, X |- (s_m∈H)∧(ζ(s_m)=0).
M |= ∃s(s∈H)∧(ζ(s)=0). This is contradiction. The non-trivial zero points of ζ(s) and the counterexample of RH
don't exist.
If |1/c|≥n for ∀n∈N,c=0 in the standard mathematics. But it may be that M |=|1/c|≤d and M |= c≠0 and ¬(X |- c=0).
The non-trivial zero points of ζ(s) are such ones.

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