My discovery about mathematics

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I prove that Riemann hypothesis(RH) isn't provable.

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Theorem 1. If the standard mathematics(SM) is consistent,RH isn't a theorem of SM.
Proof. Let X be a countable and consistent axiom system of SM. Let d be a free
individual symbol. The mathematical axioms in X don't include d. Let N" be an
individual symbol for which N"=N for the standard model of X.
X∪{d≥1,…,d≥n,1/2^d≥0,d∈N"} has a model for which d=n+1 and N"=N and is consistent for ∀n∈N.
So X∪{d≥1,d≥2,…(infinitely),1/2^d≥0,d∈N"} is consistent and has a model M.
M |= 0≤1/2^d≤1/2ⁿ for ∀n∈N.
Put t=1/2+1/2²+…(infinitely). t is a repeating binary fraction and is an
element of the real number field and t=1/2+t/2. So t=1.
1-1/2ⁿ=1/2+…+1/2ⁿ→t=1 (n→∞, n∈N).
If 1/2^d≠0, ∃m∀n((n∈N)∧(n≥m)→(1-(1-1/2ⁿ)=1/2ⁿ≤|1/2^d|)). This doesn't occur for M.
So M |= 1/2^d=0. If M |= r≥1, M |= e^-rd=(1/2^d)^{r/log 2}=0. M |= 1/cos(rdi)=0 (i²=-1) then.
You may set up the next axioms in X.
∀s((s∈C")→(ζ(s)=(2π)^sζ(1-s)/(2Γ(s)cos(πs/2)))), ∀s((s∈C")→(0/Γ(s)=0)),
∀s((s∈C")∧(s≠1)→(ζ(s)0=0)),∀s((s∈C")∧(0≤Re(s)≤1)→((2π)^s0=0)).
These are formed for the standard model for which C"=C. By these, M |= ζ(di)=0. M |= Re(di)=0
RH=∀s((s∈C")∧(ζ(s)=0)∧(0≤Re(s)≤1)→(Re(s)=1/2)).
M |= ¬RH. So ¬(X |- RH). If SM is consistent, you may set up that X is the set of
the theorems of SM. If RH is a theorem of SM, X∋RH. X |- RH then. This is contradiction.[]

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M |= 1/2^d=0=1/(2^d2). M |= 2^d=∞=2^d2=2^d+2^d ? Fermat conjecture isn't provable?
SM may not be consistent.