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.

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I prove that Riemann hypothesis (RH) isn't provable in the consistent mathematics.

                               2

Let X be a countable axiom system of the mathematics. Let d be an individual symbol which X doesn't include.
You may set up that the natural numbers are individual symbols and (n≤m)∈X and (k=n"+m")∈X when
n≤m and k=n"+m" as usual natural numbers.
Let L be a countable model of X. Let S be the object domain of L. Set up that S={c_n| n∈N}.
For the c_n in the formula in L, n∈N⊂S. If I is an infinite set in L, ∃a(L |= (a∈I)∧(I～I-{a})) by
G={(c_m,c_n)| L |= ((c_m, c_n)∈I×I)∧(m≤n-1)∧∀p((p≤m)→(c_p≠c_n))∧∀k((m+1≤k≤n-1)→(c_k∉I)∨∃h((h≤m)∧( c_k=c_h))).}
and a=c_u (u=min{n| L |= c_n∈I}). G is a set in L by the axiom of subsets.
Theorem 1. There is a model of X in which RH is false if X is consistent..
Proof. If X is consistent, X∪{ζ(d)=0, Re(d)≠1/2, d≠-2,…,d≠-2n} has a model in which
d=-2(n+1) and is consistent for ∀n∈N.
So X∪{ζ(d)=0, Re(d)≠1/2, d≠-2, d≠-4, …(infinitely)} is consistent and has a countable model M.
d in M is a non-trivial zero point of Riemann's zeta function. M |= Re(d)≠1/2. RH is false in M. ♦
Let M be M in the proof of theorem 1.
Set up that ∀I∀I"(((I∩I"=∅)→(∑_i∈I∪I" 1=(∑_i∈I 1)+(∑_i∈I" 1))∧((I～I")→(∑_i∈I 1=∑_i∈I" 1)))∈X
and ∀a(∑_i∈{a} 1=1)∧(∑_i∈∅ 1=0) ∈X . X |- ∀a∀I((a∈I)→(∑_i∈I 1=(∑_i∈I-{a} 1)+1.
If I is a finite set, (∑_i∈I 1)∈N∪{0}. If I is an infinite set, ∃a( M |=(a∈I)))∧(I～I-{a}). M |= ∑_i∈I 1=(∑_i∈I 1)+1
Let N" be an individual symbol for which ∀n((n∈N")↔∃I((n=∑_i∈I 1)∧(n≠n+1))) ∈X. N"=N∪{0} in M.
RH=∀s(((ζ(s)=0)∧(Re(s)≠1/2)→∃n((n∈N"-{0})∧(s=-2n))). X |- RH ⇒ M |= RH
But M |= ζ(d)=0)∧(Re(d)≠1/2)∧∀n((n∈N"-{0})→(d≠-2n)). X doesn't prove RH if X is consistent.
By Skolem's theorem, you cannot define the set N in the formalism. If X is consistent,
X∪{ d∈N", d≠0, d≠1,d≠2,…,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". N"=N∪{0} in M"
but M" |= d≠n for ∀n∈N∪{0}. X isn't consistent and proves RH and ¬RH.

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The mathematics which you know isn't consistent and proves every conjecture.