My discovery about mathematics

Supplement.
1 You may put 0'=(∅,∅) and 1'=({∅},{∅}).
2 If C is a set, 2^C-{∅} is a set and
X={y|∃I((I∈2^C-{∅})∧∀z((z∈y)↔∀w((w∈z)→∃s∃t((w=(s,t))∧(s∈I)∧(t∈F₂))
∧∀u∀v∀p(((u,v)∈z)∧((u,p)∈z)→(v=p))∧∀q((q∈I)→∃h((h∈F₂)∧((q,h)∈z)))))}
is a set by the axiom of replacement. R=∪X is a set then.
3 R"={y|∃x((x∈R)∧∀z((z∈y)↔∀w(((w,1')∈z)↔((w,1')∈x))))}
is a set if R is a set.
4 When u={a_i}_i∈I, v={e_i}_i∈J, w={g_i}_i∈K, (u+v)+w={d_i}_{i∈I∪J∪K}=u+(v+w).
(d_i=a_i+e_i+g_i for i∈I∩J∩K, d_i=a_i+e_i for i∈(I∩J)-K, d_i=e_i+g_i for i∈(J∩K)-I,
d_i=g_i+a_i for i∈(K∩I)-J, d_i=a_i for i∈I-(J∪K), d_i=e_i for i∈J-(K∪I), d_i=g_i for i∈K-(I∪J).)
5 For v∈R, v+{0'}_i∈I～v+{0'}_i∈J (I,J ∈2^C-{∅})
6 If u={a_i}_i∈I∈F₂^I is a set, J={x|(x∈u)∧∃y(x=(y,1')} is a set. {y|∃x((x∈J)∧(x=(y,1'))}={i|(i∈I)∧(a_i=1')} is a set
then.　Similarly, {i|(i∈I)∧(a_i=0')} is a set too.
7 If u,u"∈R and u～u", u={a_i}_i∈H, u"={a"_i}_i∈H" (H,H" ∈2^C-{∅}),
I={i|(i∈H)∧(a_i=1'}={i|(i∈H")∧(a"_i=1')} and J"={i|(i∈H)∧(a_i=0')} and K"={i|(i∈H")∧(a"_i=0') are sets
if C is a set.
If I≠∅, you can write u={1'}_i∈I+{0'}_i∈J, u"={1'}_i∈I+{0'}_i∈K. (If J"≠∅, J=J". If J"=∅,J=I. If K"≠∅,K=K". If K"=∅,K=I.)
For v∈R, v+u=(v+{1'}_i∈I)+{0'}_i∈J～(v+{1'}_i∈I)+{0'}_i∈K=v+u"
If I=∅, v+u=v+{0'}_i∈H～v+{0'}_i∈H"=v+u".
Similarly, u+v=u"+v. u～u",v～v"⇒ u+v～u"+v～u"+v".
8 Assume ¬(x-1∈N")∧(x≠1). If z=x-1, z∉N". z∉N"-{x} (z∈N"-{x})→(z∪{z}∈N"-{x}) is true.
If z≠x-1,z∪{z}≠x and (z∈N"-{x}⇒z∈N"⇒(z∈P)∧(z∈y for y for which ∀w((w∈y)→(w∪{w}∈y))∧(1∈y))
⇒(z∪{z}∈P)∧(z∪{z}∈y) ⇒ z∪{z}∈N"⇒ z∪{z}∈N"-{x}). (z∈N"-{x})→(z∪{z}∈N"-{x}) is true.
So ∀z((z∈N"-{x})→(z∪{z}∈N"-{x})) is true and 1∈N"-{x}. If z∈N",z∈N"-{x}. N"⊂N"-{x}
9 When x₁∋x₂∋…(infinitely),{x₁,x₂,…} is countable and isn't a set.
10 Nonstandard mathematics got by using x∈N" instead of N(x) may be consistent But you
cannot calculate most of elements of N" for N" is an infinite set and isn't countable.
Do such numbers have sense?
11 Let s(x,y) be a function symbol. You may set up the axioms ∀x∀y(s({x},{(x,y)})=y and
∀x∀y∀z∀w((x∉y)→(s(y∪{x},z∪{(x,w)})=s(y,z)+w)). s(I,{x_i}_i∈I)=∑_i∈Ix_i then.