My discovery about mathematics

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I show a paradox about the zeta function. You can prove and can deny BSD conjecture if L(E,1)=0.
Riemann hypothesis is nonsense too.

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Put F=Z[[X]],O=Z[X]. Define A+B={a+b|a∈A,b∈B} for A,B∈F/O.
O is the zero of the module F/O.
Theorem 1. O+O+…(infinitely)≠O
Proof. O+O+…(infinitely)={c₁+c₂+…(infinitrly)|c₁∈O,c₂∈O,…}
∋1+X+X²+…(infinitely)∉O ♦
Put D={x|(x∈C)∧(Re(x)≥2)},S={x|(x∈C)∧((Im(x)=0)∨(0≤Re(x)≤1))}.
Let p₁=2,p₂=3,…p_m,… be the prime numbers.
Let g(p,t) be a meromorphic function about t on C for which g(p,t)≠0 for ∀t∈C.
Let f(P,t) be a meromorphic function about t on C for which ∀x((x∈D)→(f(P,x)=∏_p∈Pg(p,x))).
Theorem 2. Theory of ζ(x) isn't consistent.
Proof. Put g(p,t)=1/(1-1/p^t),
V={h(t)|(h(t) is a meromorphic function on C)∧∀x((x∈C)∧(h(x)=0)→(x∈S))}.
Define s(h(t))=O for∀h(t)∈V.
f({2},t)=g(2,t)≠0 for ∀t∈C. f({2},t)∈V s(f({2},t))=O
f({2}∪{3},t)=g(2,t)g(3,t)≠0 for ∀t∈C.
f({2}∪{3},t)∈V s(f({2}∪{3},t))=O=O+O=s(f({2},t))+s(f({3},t))
f({p₁}∪…∪{p_m},t)=g(p₁,t)…g(p_m,t)≠0 for∀t∈C. f({p₁}∪…∪{p_m},t)∈V
s(f({p₁}∪…∪{p_m},t))=O=O+…+O=s(f({p₁},t))+…+s(f({p_m},t)),…
Repeating this infinitely, you get
s(ζ(t))=s(f({p₁}∪{p₂}∪…(infinitely),t))=s(f({p₁},t))+s(f({p₂},t))+…(infinitely)
=O+O+…(infinitely)≠O ζ(t)∉V ∃x((x∈C)∧(ζ(x)=0)∧(x∉S))
This is contradiction by the known theorem.♦ Riemann hypothesis is nonsense.
Theorem 3. You can prove and can deny BSD conjecture if L(E,1)=0.
Proof. assume that L(E,1)=0. You can set up that f({p₁}∪{p₂}∪…(infinitely),t)=L(E,t)
by a suitable g(p,t).
Put V"={h(t)|(h(t) is a meromorphic function on C)∧∀m((m∈N)∧(t=1 is a zero point
of h(t) whose order is m.)→q(m))}, s"(h(t))=O for ∀h(t)∈V".
f({p₁}∪…∪{p_m},1)=g(p₁,1)…g(p_m,1)≠0
f({p₁}∪…∪{p_m},t)∈V" s"(f({p₁}∪…∪{p_m},t)=O=O+…+O=f({p₁},t)+…+f({p_m},t)
So s"(L(E,t))=s"(f({p₁}∪{p₂∪…(infinitely),t))=s"(f({p₁},t))+s"(f({p₂},t))+…(infinitely)
=O+O+…(infinitely)≠O. L(E,t)∉V" ∃m((m∈N)∧(t=1 is a zero point of L(E,t)
whose order is m.)∧(¬q(m))
If you put q(m)=(m≠rank(E(Q))), BSD conjecture is proved.
If you put q(m)=(m=rank(E(Q))), BSD conjecture is denied.♦
Theory of L(E,t) isn't consistent.

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The above means that you must not use infinite symbols like {p₁}∪{p₂}∪…(infinitely)
to express functions. One which isn't consistent isn't only theory of zeta functions but
the standard mathematics. Actually,N={1}∪{2}∪… and {p₁}∪{p₂}∪… aren't sets.
I have written about it in another paper. The physics isn't consistent too.
Every phenomenon like the free productivity is possible. This is an answer to the 6th
problem of Hilbert.