My discovery about mathematics

Fermat conjecture(FC) and Beal conjecture(BC) and Riemann hypothesis(RH) aren't provable in the consistent
mathematics based on the set theory. FC is proved. The set theory may not be consistent.
Let X be a countable axiom system of the set theory. Let's write ∅=0, x∪{x}=S(x), S(0)=1,S(1)=2,…,
N={0,1,2,…}. (0∈I)∧∀x((x∈I)→(S(x)∈I)) ∈X. ∀x((x∈N")↔(x∈I)∧∀y((0∈y)∧∀z((z∈y)→(S(z)∈y))→(x∈y))) ∈X.
Let d be an individual symbol which X doesn't include.
If X is consistent, X∪{ d∈N",d∋0,d∋1,…d∋n} has a model in which N"=N and d=n+1 and is consistent.
So X∪{d∈N", d∋0, d∋1, d∋2,…(infinitely)} is consistent and has a countable model M. d∈N"-N in M.
If N is a set in M, N"-N=∅. So N isn't a set in M.
You are to define the next functions in M. M |= (m+0=m)∧(m+S(n)=S(m+n). M |= (m0=0)∧(mS(n)=mn+m).
M |= (m⁰=1)∧(m^S(n)=mⁿm). The axioms of subsets in X don't include the function symbols x+y,xy and x^y.
You can define M |= mⁿ=0 when n∈N"-N in M. S(n)∈N"-N in M. ⇔ n∈N"-N in M.
M |= m^S(n)=0=0m=mⁿm is true. M |= ∅=0=2^d～∏_i∈d 2. The axiom of choice isn't true in M.
This M is a model of X"=X∪{∀m∀n((m∈N")∧(n∈N")→(m+0=m)∧(m+S(n)=S(m+n))∧…∧(m⁰=1)∧(m^S(n)=mⁿm)}
When n≠0, {m | n^m≠0}=N∪{0} isn't a set in M. M |= n^m≠0 isn't proved by induction.
H={n | n∈N")∧W(n)} may not be a set. You cannot use the induction in the set theory.
Put FC=∀x∀y∀z∀n((x∈N"-{0})∧(y∈N"-{0})∧(z∈N"-{0})∧(n∈N"-{0,1,2})→(xⁿ+yⁿ≠zⁿ)).
M |= (2∈N"-{0})∧(d∈N"-{0,1,2})∧(2^d+2^d=0+0=0=2^d). M |= ¬FC. X" doesn't prove FC.
M |= 2^d+3^d+1=0+0=0=5^d+2. M |=¬BC. X" doesn't prove BC.
Put Y=X"∪(Z"={n-m | (n-m⊂N"×N")∧∀x∀y((( (x,y)∈(n-m))↔(n+y=m+x))})
∧∀n∀m∀n"∀m"(((n-m)+(n"-m")=n+n"-(m+m"))∧((n-m)(n"-m")=(nn"+mm"-(nm"+mn"))∧
((n-m)-(n"-m")=n+m"-(m+n")))∧(Q"={u/v | (u/v⊂Z"×(Z"-{0-0}))∧∀x∀y(((x,y)∈u/v)↔(xv=uy))})∧
∀u∀v∀u"∀v"(u/v+u"/v"=(uv"+u"v)/vv")∧((u/v)(u"/v")=uu"/vv")∧((u/v≥u"/v")↔∃n((n∈N")∧(uvv"²-u"v"v²=n-0))
∧(R"={ [a,b] | (a∪b=Q")∧(a∩b=∅)∧(a≠∅)∧(b≠∅)∧∀x∀y((x∈a)∧(y∈b)→(x≤y)∧
(a={r |(r∈Q")∧P(r)∧(P(x) is an arithemetical logical formula.)})})∧
([a,b]+[a",b"]=[{x+x" | (x∈a)∧(x"∈a"), {y+y" | (y∈b)∧(y"∈b")}])
∧(C"=R"×R")∧∀x∀y(Re((x,y)=x)∧Im((x,y)=y)∧((x,y)+(x",y")=(x+x",y+y"))∧((x,y)(x",y")=(xx"-yy", xy"+yy").
Y is the definition of C",R",Q" and Z" by N" and axioms of subsets etc..
You can make a model of Y from M. Let M" be it.
M" |= ∀n((n∈N")→(n^d=0)). Let P be the set of the prime numbers in M". P⊂N" in M"
M" |= ζ(d)=∏_p∈P(p^d-0)/(p^d-1))=∏_p∈P(0-0)/(0-1)=0. M" |= Re(d)≠1/2. M" |= Re(d)≥2. M" |= ¬RH
Y doesn't prove RH.
Theorem 1 X isn't consistent.
Proof. Assume that X is consistent. M is a model of X". M" is a model of Y. If n∈N in M, M" |= 2ⁿ≤d for 2ⁿ∈N
in M. If n∈N"-N in M, M" |= 2ⁿ=0≤d.
Put a"={r | (r∈Q")∧∃n((n∈N")∧(r≤2ⁿ)}. M" |= 1∈a"≠∅. M" |= d(=(d-0)/(1-0))∈Q"-a"≠∅.
M" |= S=[a",Q"-a"]∈R". ∃n∃n"((n∈N")∧(n"∈N")∧(r≤2ⁿ)∧(r"≤2^n")→(r+r"≤2^m+1)∧(2^m=max{2ⁿ, 2^n"})).
M" |= S+S=S. R" is a module. M" |= -S=[u,Q"-u]. (u={-u"| u"∈Q"-a"}.) M" |= 1≤S=0 This is contradiction.♦
Y isn't consistent and proves FC,¬FC, BC, ¬BC, RH, ¬RH. Actually,FC is proved.