My discovery about mathematics

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Let X be a countable axiom system of the mathematics. You may set up that
Y={a≠b, a=a}⊂X and {R(t)| Y |- t≠a}⊂X and
{∀x∀y((x=x)∧((x=y)→(P(x)↔P(y)))) | P(z) is the logical formula.}⊂X.
a and b are individual symbols. R(x) is a predicate symbol.
Theorem 1. X isn't consistent.
Proof. Assume that X is consistent. Let d be an individual symbol which X doesn't include.
If X∪{d=b} isn't consistent, X |- d≠b. X |- ∀x(x≠b) X |- b≠b. X isn't consistent.
So X∪{d=b} is consistent and has a model M. M is a model of X too.
X |- (d=b)→(R(d)↔R(b)). M |= (d=b)→(R(d)↔R(b)). (M|= (d=b))⇒(M |= (R(d)↔R(b))).
M |= (d=b) is true. So M |= (R(d)↔R(b)) is true. (M |= R(d))⇔ (M |= R(b)) is true. M |= R(b)(⇔ Y |- b≠a) is true.
So M |= R(d) is true. Y |- d≠a ⇒ Y |- ∀x(x≠a) ⇒ Y |- a≠a. Y |- a=a. Y |- a≠a is false for Y(⊂X) is consistent.
So Y |- d≠a is false. This is contradiction.♦
Put R"(s)=(ζ(s)=0)∧(0≤Re(s)≤1)→(Re(s)=1/2).
If X∪{R"(d)} isn't consistent, X |- ¬R"(d). X |- ∀x(¬R"(x)). X |- ¬R"(2). Tis is false. X |- RH ⇔ X |- R"(d).

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