My discovery about mathematics

Naoto Meguro. Amateur.
MSC2010. Primary 03C62; Secondary 03C55.
Key Words and Phrases. The natural number theory in the set theory,
Gödel's theorem.
The abstract. The natural number theory in the set theory isn't consistent.

                              1

I indicate that the set theory(ST) including the natural number theory(NNT) isn't consistent
by Gödel's theorem.

                              2

Let's write ∅=0 and x∪{x}=x+1. Let X be a countable and consistent axiom system of
ST including NNT. Let d be a free individual symbol. X doesn't include d. Let N" be an individual
symbol for which (1∈N")∧∀n((n∈N")→(n+1∈N")) ∈X and
∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)) ∈X.
You may set up that ∀x(x≠1+x) ∈X and ∀n((n∈N")→(n=∑_m∈n1^m)) ∈X and
∀x∀y((x∈y)→(∑_m∈y1^m=1^x+∑_m∈y-{x}1^m))∈X
for these are formed in the case of N"=N.
Theorem 1. X has a countable model M for which M |= (d∈N")∧(m∈d) for ∀m∈{0}∪N.
Proof. The axiom system X∪{d∈N",0∈d,1∈d,…,m∈d} has a model M_m for which
M_m |= N"=N and M_m |=d={0,1,…m}=m+1 and is consistent for ∀m∈N.
So X∪{d∈N",0∈d,1∈d,2∈d,…(infinitely)} is consistent and has a countable model M.
M is a model of X and M |= (d∈N")∧(m∈d) for ∀m∈{0}∪N.♦
Theorem 2. X doesn't exist.
Proof.Assume existence of X. For M in theorem 1, d becomes a countable set.
You can put M |= d={n₁,n₂,…(infinitely)}∈N". So M |= d=∑_m∈d1^m=1^n₁+1^n₂+…(infinitely).
M |= 1+d=1+(1+1+…(infinitely))=1+1+1+…(infinitely)=d. But M |= 1+d≠d.This is contradiction.♦

                              3