My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary 03C65; Secondary 30D30.
Key Words and Phrases. Non-standard model, Riemann hypothesis.
The abstract. Riemann hypothesis isn't provable.

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I prove that Riemann hypothesis(RH) isn't provable in the standard mathematics.

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Let X be the countable axiom system of the standard mathematics. Let d be a free individual symbol.
X doesn't include d. Let f(x) and z(x) be free function symbols. X doesn't include f(x) and z(x).
Set up that X includes the functions (2π)^xi/Γ(xi) and ζ(x) and the theorems of them.
Set up that X∋ RH=∀s((s∈C)∧(Re(s)≠1/2)∧(Re(s)≥0)→(ζ(s)≠0)).
Let N and C be the individual symbols which become N and C in the standard model of X.
Theorem 1. If X is consistent, Y=X∪{d∈N, d≥1,d≥2,d≥3,…(infinitely)} is consistent.
Proof. X∪{d∈N, d≥1,d≥2,…,d≥m} has a model in which d=m+1 and N=N and C=C and
is consistent for ∀m∈N. So Y is consisent.♦
Assume that X is consistent.
Let M be a countable model of Y. Let D be the object domain of M. s∈D⇒ζ(s)∈D etc..
Put Z={f(1)=e^π/2, ∀n(f(n+1)=f(n)e^π/2), ∀n(z(ni)=ζ(1-ni)(2π)ⁿⁱ/(Γ(ni)(f(n)+1/f(n)))}.
Define M" by M" |= P when M |= P and M" |= f(n)=e^πn/2 when n∈C∩D and M" |= f(n)=0 when n∈D-C
.and M" |= z(s)=ζ(s) when s∈C∩D and; M" |= z(s)=0 when s∈D-C. The object domain of M" is D.
M" is a model of Y∪Z or X∪Z.. M" |= ∀s(ζ(s)=0)→(z(s)=0)
You may set up that ∀n((n∈N)→((ζ(1-ni)(2π)ⁿⁱ/Γ(ni))0=0)∧(ni∈C)∧(Re(ni)≠1/2)∧(Im(ni)≥0)) ∈X for this is true in
the standard mathematics in which N=N and C=C.
Theorem 2. X isn't consistent.
Proof. Assume that X is consistent. Y∪Z |- RH, M" |= ∀s((s∈C)∧(Re(s)≠1/2)∧(Re(s)≥0)→(ζ(s)≠0)).
But M" |= (d∈N)∧(z(di)=ζ(1-di)(2π)^di/(Γ(di)(f(d)+1/f(d)))=0). This is contradiction.♦
In the standard model of X∪Z in which N=N and C=C and f(x)=e^πx/2 and z(x)=ζ(x),.
RH isn't provable in the standard mathematics.

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If someone proves RH from X∪Z, the standard mathematics isn't consisent.