My discovery about mathematics

Naoto Meguro . Amateur. 08A65.

                               1

You cannot prove Riemann hypothesis in the consistent mathematics.

                                2

Let X be an axiom system. Let c be an individual symbol which X doesn't include.
Put T"={x|(x∈C)∧(ζ(-2x)=0)∧(Re(-2x)≠2^{- 1})}={s₁,s₂,…}.
Set up that (s(n)=s_n)∈X for ∀n∈N.
Let S be the set. Let (S) be the individual symbol which becomes S in the standard
mathematics. Set up that n∈S ⇒ (n∈(S))∈X. Let's write N=(N) and T=(T").
Add the rule of inference X |-" P(n) for ∀n∈S⇒ X |- " ∀x((x∈(S))→P(x)).
Set up that X |- P ⇒ X |-" P. Define that X is S-consistent ⇔ ¬(X |-" P∧(¬ P)).
Put Y= X∪ (∪_n∈T"{c≠n}). Put q(x)=(x∉T). X |-" (c=n)→(¬q(c)) for ∀n∈T".
Theorem 1. Y |-" P ⇒ X |-" q(c)→P.
proof. X |-" q(c)→(c≠n) for ∀n∈T". So it is enough to think the case that
Y|-" ∀x((x∈(S))→P(x)) is proved by Y |-" P(n) for ∀n∈S. By induction,
X |-" q(c)→P(n) . X |-" ∀x((x∈(S))→(q(c)→P(x))). X |-" ∀x(q(c)→((x∈(S))→P(x))).
X |-" ∀xq(c)→∀x((x∈(S))→P(x))). X |-" q(c)→∀x((x∈(S))→P(x)).♦
Theorem 2. X |-" P(c) ⇒ X |-" ∀xP(x).
Proof. It is enough to think the case that
X |-" ∀x((x∈(S))→P(x,c) is proved by X |-" P(n,c) for ∀n∈S.
By induction,X |-" ∀yP(n,y), X |-" ∀x((x∈(S))→∀yP(x,y)). X |-" ∀y∀x((x∈(S))→P(x,y)).♦
Theorem 3. Y |-" c∉N and Y |- c∉T.
Proof. Y |-" c≠n for ∀n∈N⊂T". Y |-" ∀x((x∈N)→(c≠x)) Y |-" (c∈N)→(c≠c).
Y |-" (c=c)→(c∉N). Y |-" c∉N. Similarly, Y |- c∉T♦
Theorem 4. Y is S-consistent if X is S-consistent and X |- G.
Proof. Y |-" 1≠1 ⇒ X |-" q(c)→(1≠1). X |-" (1=1)→(¬q(c)). X |-" ¬q(c). X |-"∀x(¬q(x)).
X |-" ¬q(- 1). X |- - 1∈T. X isn't consistent.♦
Theorem 5.
proof. Y |- " c∉T. ♦

                                     3

Substituting Γ(x)^{- 1} for ζ(x), you see that the complex function theory isn't N-consistent
and proves every conjecture and its denial by the above plausible rule of inference.