My discovery about mathematics

Naoto Meguro. Amateur.
MSC2010. Primary;SEcondary.
Key Words and Phrases. Goldbach conjecture,non-standard model.
The abstract. No contradiction is led even if you deny Goldbach conjecture.

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Let P={p₁=2,p₂=3,…,p_m,…(infinitely)} be the set of the standard prime numbers.
Put X={∀x∀y(((S(x)=S(y))→(x=y))∧((x≠1)→∃z(x=S(z)))∧(S(x)≠1)∧(x+1=S(x))∧(x+S(y)=S(x+y))
∧(x1=x)∧(xS(y)=xy+x)∧(x¹=x)∧(x^S(y)=x^yx)∧((x|y)↔∃z(y=xz))}∪
{p(1)∧∀n(p(n)→p(S(n)))→∀mp(m)| p(t) is an arithmetical logical formula.}.
X is an axiom system of the natural number theory. The terms mean natural numbers.
Let d be a free individual symbol. X doesn't include d.
Put Y=X ∪{2|d, 3|d,…,p_m|d,…(infinitely)}=X∪{p|d | p∈P}.
Theorem 1. If X is consistent,Y is consistent.
Proof. Assume that X is consistent. X doesn't include d and has a model.
You may set up that d=p₁…p_m and get a model of Y_m=X∪{2|d,…,p_m|d}.
Y_m is consistent for ∀m∈N.
If Y |- P∧¬P, ∃m((m∈N)∧(Y_m |- P∧¬P). So Y is consistent.♦
Put GC=∀x((2|x)→∃p∃q((p∈P)∧(q∈P)∧(x=p+q))).
Theorem 2. If X is consistent, GC isn't true in the mathematics by Y.
Proof. Assume that GC is true in the mathematics by Y.
2|d. So d=p+q (p∈P, q∈P) p|d. p| (d-p)=q. So p=q . Let r be a prime number for which ¬(r|2p).
r|d=p+q=2p.This is contradiction.♦

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