My discovery about mathematics

                      1

I try to prove that the standard mathematics isn't consistent by Gödel's completeness
theorem. By the theorem,an infinite series becomes "a polynomial" for a model of the standard
mathematics. It leads contradiction.

                      2

Put k={0",1"}≅Z/2Z.
Let A be a countable axiom system of the standard mathematics.
Let T and U be individual symbols expressing variables of polynomials.
Let N be an individual symbol which expresses the set of the natural numbers for the
standard model. (1∈N)∧∀x((x∈N)→(x+1∈N)) ∈A,
∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N⊂x)) ∈A
Let P be an individual symbol for which
∀x((x∈P)↔((x⊂N∪{0})∧∃y((y∈N∪{0})∧∀z((z∈x)→(z≤y))))) ∈A.
Let d be a free individual symbol. The axioms except the logical axioms and
the axioms about equality don't include d. You may set up that
0"≠1" ∈A, T=1"+"U ∈A, k[T]=k[U] ∈A.
Theorem 1. The standard mathematics isn't consistent.
Proof. Assume consistency of A. Let's identify {m,n,…}(∈P) and T^m+"Tⁿ+"…(∈k[T]).
Set up that k[T]=P ∈A, ∀x((x∈N∪{0})→(T^x={x})) ∈A,
∀x∀y((x∈P)∧(y∈P)→(x+"y={z | (z∈N∪{0})∧(((z∈x)∧(z∉y))∨((z∉x)∧(z∈y)))})) ∈A,
∀x∀y((x∈N∪{0})∧(y∈P)→T^xy={z |(z∈N∪{0})∧∃w((w∈y)∧(z=w+x))} ∈A
∀x∀y∀z((x∈P)∧(y∈P)∧(z∈P)→((x+"y)z=xz+"yz)) ∈A.
You can evolve theory of k[T] for the standard model by these.
Put A_m=A∪{ d∈k[T], ∃x((x∈k[T])∧(d=1"+"T+"…+"T^m+"T^m+1x)) } (m∈N).
∪_m≤n A_m has a model M_n for which M_n |= d=1"+"T+"…+"Tⁿ and is consistent.
So ∪_n∈N A_n is consistent and has a countable model M for which
M |= d∈k[T], M |= d=1"+"T+"…+"T^m+1c"_m, M |= c"_m∈k[T] (m∈N).
d={0,1,…,m, elements which are larger than m } for ∀m(∈N) for M.
d={0,1,2,…,nonstandard elements} for M. d is an infinite set for a countable model M.
You can set up that M |= c∈d ⇔ (M |= c=m₁)∨(M |= c=m₂)∨…(infinitely).
(m_i≠m_k for i≠k).
d={m₁,m₂,…}, d+"T^m₁={m₂,m₃,…}, d+"T^m₁+"T^m₂={m₃,m₄…},…
d+"T^m₁+"T^m₂+"…(infinitely)={ }=∅ for M.
Put d"=T^m₁+"T^m₂+"…(infinitely).　+"d" expresses repeating infinite additions. d+"d"=∅ for M.
It is easy to see M |= c∈P ⇒ c+"d"=c+"d ∈P for M.
You may set up that ∀x((x∈k[U])↔∃1y∃1z((x=y+"Uz)∧(y∈k)∧(z∈k[U]))) ∈A,
∀x((x∈N∪{0})→∃y((y∈k[U])∧(1"+"U)^x=1"+"Uy))) ∈A for these are formed for the standard
model.
M |= T^m_i=(1"+"U)^m_i=1"+"Uc_i, M |= c_i∈k[U] (i∈N) For M, d"=(1"+"Uc₁)+"(1"+"Uc₂)+"…(infinitely)=1"+"1"+"…(infinitely)+"U(….
d"=1"+"d" for M. ∅=d+"d"=d+"1"+"d"=(d+"1")+"d"=(d+"1")+"d=1"={0} for M.
This is contradiction.♦

                      3

If this theorem isn't wrong,the final problem is to prove a contradictory proposition from A
concretely.
You are to prove every logical formula as a theorem then. If mathematics isn't consistent,
physics isn't consistent too. Every phenomenon like the free productivity is possible.
Or you must not use k[T]. Remark that you cannot define the set of the polynomials
strictly.