My discovery about mathematics

Naoto Meguro. Amateur. MSC2020 03C62.

                              1

You cannot prove Riemann hypothesis(RH) in the consistent mathematics.
It reveals defect of the set theory or the predicate logic.

                              2

Let T be an individual symbol. Let F be the set of the closed terms. Let G be the set of
the closed terms which don't include T.
Put S"={s|(s∈C)∧(ζ(-2s)=0)∧(Re(-2s)≠2^{- 1})}. RH means S"=N.
You are to treat S" and the elements of S" as individual symbols. S"∈G N⊂S"⊂G
Put U={s|(s∈F)∧(s∉S")} and Y={∃x(x∈T)}∪(∪_s∈U{s∉T}) and X=Y∪(∪_n∈N{n∉T}).
Let P be the proposition "There is a consistent set theory whose object domain is G.".
Theorem 1. X is a consistent axiom system if P is true.
Proof. Assume that P is true. By the definition T =S"-{1,2,…,n}, you get a consistent set
theory in which (s∈T⇔ s∈S"-{1,2,…,n}) and the object domain is F.
In this theory, s∉(S"-{1,2,…,n}) ⇒ s∉T. So 1∉T,…,n∉T.
s∈U ⇒ s∉S" ⇒ s∉S"-{1,2,…,n} ⇒ s∉T. n+1∈(S"-{1,2,…,n}). So n+1∈T. ∃x(x∈T).
The elements of X(n)=Y∪{1∉T, 2∉T,…, n∉T } are true and p∧(¬p) is false in this theory.
In this theory, the logical formulas proved by only logical axioms become true when you
substitute elements of F for the free variables. So ¬(X(n) |- p∧(¬p)) for ∀n∈N.
X is consistent.♦
Let p(t) be a predicate symbol. Let a be a free variable. Define p(t)=(X |- t∉T).
Let H be the set of the theorems of the standard theory whose object domain is F.
Put q=(X |- ∃x(x∈T))⇔ (X |- ¬∀x(x∉T)).
Theorem 2. RH is false if P is true.
Proof. Assume P. X is consistent. p(a)=(X |- a∉T)⇔(X |- ∀x(x∉T))⇒ ¬q. p(a)→(¬q) ∈H.
∀xp(x)→p(a) ∈H. ∀xp(x)→(¬q) ∈H. q→(¬∀xp(x)) ∈H. q∈H. ¬∀xp(x)∈H. ∃x(¬p(x))∈H.
If s∈F, s∉S"⇒ s∈U ⇒ p(s) and s∈N ⇒ p(s). ¬p(s)⇒ s∈S"-N then. ∃x(¬p(x))∈H⇒ S"-N≠∅.
RH is false.♦
The standard theory assumes that P is true. So RH is false in it if it is consistent.

                                3

If P is true, p(a)→(¬q). q→(¬p(a)). q→∀x(¬p(x)). ∀x(¬p(x)). ¬p(1). p(1).
The standard theory isn't consistent and proves every conjecture and its denial.
You may have to treat only finite sets as Gauss insisted. The above will be ghastly for him.
You couldn't define Γ(s) and ζ(s) then.