My discovery about mathematics

Naoto Meguro. Amateur
MSC 2010. Primary 03C62;Secondary 30D30.
Key Words and Phrases. Riemann hypothesis, an axiom system.
The abstract. No contradiction is led even if you deny Riemann hypothesis.

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I indicate that no contradiction is led even if you deny Riemann hypothesis(RH).

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Let X be a consistent and countable axiom system of the complex number theory.
The terms express only complex numbers. ∀x((x=Re(x)+iIm(x))∧(Re(x)²≥0)∧(Im(x)²≥0)∧(i²=-1))∈X.
∀x∀y((x>y)∨(y>x)→(Im(x)=0)∧(Im(y)=0)) ∈X.
Let d be a free individual symbol. X doesn't include d.
Set up that ζ(b₁)=c₁ ∈X, ζ(b₂)=c₂ ∈X,…{b₁,b₂,…}=Q(i),
The c_k is the standard value of Riemann's zeta function ζ(s) at s=b_k. ζ(-2)=0 ∈X,ζ(-4)=0 ∈X,…
And ∀x((x≠1)→∀h((h≠0)→∃u((u≠0)∧∀y((|x-y|≤|u|)→(|ζ(x)-ζ(y)|≤|h|))))) ∈X.
Put Y=X∪{ζ(d)=0,Re(d)≠1/2,d≠-2,d≠-4,d≠-6,…(infinitely)}.
Theorem 1. Y is consistent.
Proof. X has a model and doesn't include d. You may set up that d=-2(m+1)(m∈N) and get a model of
Y_m=X∪{ζ(d)=0,Re(d)≠1/2,d≠-2,d≠-4,…,d≠-2m}. Y_m is consistent for ∀m∈N.
If Y |- p∧¬p, ∃m((m∈N)∧(Y_m |- p∧¬p)). So Y is consistent.♦
The d in Y becomes a counter-example of RH=∀s((ζ(s)=0)∧(Re(s)≠1/2)→(s=-2)∨(s=-4)∨…).
You can think the consistent mathematics by Y in which Y |- (ζ(d)=0)∧(Re(d)≠1/2)∧(d≠-2m) for ∀m∈N
and ¬RH(⇔∃s((ζ(s)=0)∧(Re(s)≠1/2)∧(s≠-2)∧(s≠-4)∧…)) is true.

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The consistent Y is gotten by adding ¬RH by countable logical formulas to X. No contradiction is led
even if you assume ¬RH. RH resembles the axiom of parallels or Continuum Hypothesis.