My discovery about mathematics

Theorem 5.　You cannot evolve theory of the direct sum by any countable axiom system.
Proof. Assume that A evolve theory of the direct sum.
Put p(x)=∏_i∈x A(i) , s(x)=⊕_i∈x A(i) , A(i)=Z/2Z={0",1"}=k, p_i: the canonical projection.
(i∈x),
E_m=A∪{m∈d, d∈N, p(d)=s(d)} (m∈N).
Like the proof of theorem 1,you get that ∪_m∈N E_m is consistent and has a countable model M
for which M |= m∈d for ∀m∈N and M |= p(d)=s(d) and M |= d∈N.　d is an infinite set for M.
You may set up that
∀x∃y((y∈p(x)=k^x)∧∀z((z∈x)→(y:x∋z→1"∈k))) ∈A,
∀x∀y((x∈N)∧(y∈s(x))→((∑_i∈x p_i(y))∈k)∈A,
∀x∀y∀z((x∈y)∧(z∈s(y))→(∑_i∈yp_i(z)=p_x(z)+∑_i∈(y-{x})p_i(z)))∈A,
∀x(∑_i∈Øp_i(x)=0")∈A
for these are formed for the standard model.
∃c((M |= c∈p(d)) ∧ (M |= c=(1",1",1"…(infinitely)…)))
You can set up that d={c₁,c₂,…} for M for M is a countable model.
M|= c∈s(d) M |= d∈N
∴ M |= e=∑_i∈d p_i(c)=p_c1(c)+p_c2(c)+…(infinitely)…=1"+1"+…(infinitely)…∈k
You need infinite terms to decide e.
Seeing the sequence of symbols,you get M |= e=1"+e. M |= 0"=e+e=1"+e+e=1"+0"=1"
This is contradiction. A isn't consistent.
I have used logic of theory of the direct sum which you may not be able to write by
logical formulas. Such theory proves that A isn't consistent. If A is consistent,
such theory isn't consistent.♦