My discovery about mathematics

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

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Let X be a countable axiom system of the standard mathematics. Let d be a free individual symbol.
X doesn't include d. Let f(n.s) be a free function symbol. X doesn't include f(n.s).
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 too.
Proof. X∪{d∈N, d≥1,d≥2,…,d≥m} has a model in which d=m and N=N and C=C
and is consistent for ∀m∈N. So Y is consistent.♦
Let M be a countable model of Y. Let D be the object domain of M. Put p=∀n∀s(nf(n,s+1)=f(n,s)).
Define M" by M" |= P when M |= P and M" |= f(n.s)=1/n^s when n,s∈C∩D and M" |= f(n,s)=0 when s∈D-C
or n∈D-C.
M" is a model of X∪{p} and Y∪{p}. Put RH"=∀s((s∈C)∧(Re(s)>1)→(∑_n∈Nf(n,s)≠0)). RH⇒RH".
Theorem 2. If X is consistent, ¬(Y∪{p} |- RH")
Proof. If Y∪{p} |- RH", M" |= RH".
M" |= ∑_n∈Nf(n,di+2)=∑_n∈N 0=0, M"|= di+2∈C, M" |=Re(di+2)=2>1. M" |= ¬RH". This is contradiction.♦

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