My discovery about mathematics

Let Z be an individual symbol which expresses the set of the integers
for the standard model.
r∈Z ⇔ r∈N or r=-m,m∈N.
The next theorems may be wrong but I submit them as ideas.
Theorem 6. If f(x)∈Z[x],there is a model M for which
M |= ∃x((x∈Z)∧(f(x)=0)) for A if A is consistent.
Proof. Put C(x)=x+f(x) for f(x)=r₀+r₁x+…+r_sx^s.
You may set up that
∀x∀y(C({x},y)=C(y))∈A.
Assume that C({x,y,…},z)=C({x},C({y},C(…(z)….
This is definition of the function C(x,y) for the countable models.
∀x((x∈N)→(C(x,1)∈Z))∈A,
∀x∀y((x∈N)∧(y∈N)→(x∈y)∨(x=y)∨(y∈x))∈A.
These are formed for the standard model.
Let M be a model of A for which N isn't a set.
If k∈ N-N, 1∈2∈…,…∈k-1∈k. k is an infinite set.
C(k,x)=C(C(…(infinitely)…(1))…) for M.
Seeing the sequence of symbols,you get M |= C(k,1)=C(C(k,1)). M |= f(C(k,1))=0
M |= C(k,1)∈Z ♦
Theorem 7 if f(x)∈Z[x], ∃x((x∈Z)∧(f(x)=0)) in formalism.
Proof. Put P=∃x((x∈Z)∧(f(x)=0)). Set up that ¬P∈A. If this A is consistent,
∃M,M is a model of A and M |= P by theorem 4. But M |= ¬P by ¬P∈A.
A isn't consistent. Put A"=A-{¬P}. A" |- P　A" is an axiom system of the standard
mathematics too.♦